The hypotenuse of a triangle is 10 inches long. One of the legs is 6.5 inches long. what is the measurement of the missing leg? round to the nearest tenth, if necessary

Answer:

-9/14

Step-by-step explanation:

Hey there!

Guide:

• Difference means subtract/subtraction
• Product means multiply/multiplication
• Sum means add/addition
• Quotient means divide/division

• Now that we know what “product” means… we can make the question/equation easier to solve.

3/4 × -6/7

= 3(-6) / 4(7)

= -18 / 28

= -18 ÷ 2 / 28 ÷ 2

= -9 / 14

Therefore, your answer is: -9/14

Good luck on assignment & enjoy your day!

~Amphitrite1040:)

Step-by-step explanation:

2× (4x-5y) = 21

3x+10y= -53

-------------------------

8x-10y = 42

3x+10y=-53

------------------------

11x= -11

{x=-1}

-----------------

3x+10y=-53

x=-1

3(-1)+10y=-53

-3+10y=-53

10y= -50

{y=5}

Answer:

7x - 3, x being the number

Step-by-step explanation:

number = x

7x - 3

Answer:

Three less than 7 times a number is 39.

Step-by-step explanation:

Answer:

  A. Find h(10) = 0 . . . (it is not)

  B. h(5.235) = 0; hang time was about 5.235 seconds

Step-by-step explanation:

The hang time of the ball is the value of x in h(x) such that h(x) = 0.

__

Juan can show the hang time is 10 seconds by finding h(10) = 0. The value of h(10) is ...

  h(10) = -16(10²) +83(10) +4 = -1600 +834 = -766 ≠ 0

Juan cannot prove the hang time was 10 seconds.

__

The hang time is the solution to ...

  h(x) = 0

  -16x² +83x +4 = 0 . . . . . use the given expression for h(x)

  x² -5.1875x = 0.25 . . . . divide by -16

  x² -5.1875x +6.7275390625 = 6.9775390625 . . . . add (5.1875/2)²

  (x -2.59375)² = 6.9775390625 . . . . . . . . . . rewrite as a square

  x = 2.59375 ±√6.9775390625 ≈ {-0.048, 5.235} . . . . square root, find x

The hang time of the ball was about 5.235 seconds.

__

We like a graphing calculator for finding a quick solution to questions like this.

✿————✦————✿————✦————✿————✦————✿

So, to find the mean they added up the scores from 20 seasons and divided by 20, the number of seasons to get an average score of 10.4.

10.4 = (x1 +x2 +x3 ... +x20 )/20

Then;

10.4*(20) = (x1 +x2 +x3 ... +x20 )

208 = sum of the first 20 seasons

We want to add one more number in there, the 21st season with a score of 14. Add the 14 to the sum of the first 20 seasons and then divide by 21, the new number of seasons you're averaging.

Mean = (208+14)/21  

✿————✦————✿————✦————✿————✦————✿

Answer

= 10.6

✿————✦————✿————✦————✿————✦————✿

#carryonlearning

Answer:

7.1 x 10³

Step-by-step explanation:

Move your decimal so that it follows your first whole number greater than zero. The number of places you moved the decimal is the same as the exponent. Since the number 7.1 needs to INCREASE to be 7,100, then the exponent is positive.

Answer:

7.1 x 10[tex]^{3}[/tex] (note the 3 is an exponent)

Step-by-step explanation:

1. Count how many places there are to the right of 7

2. There are 3

3. Apply the formula of scientific notation a x 10[tex]^{b}[/tex]

4. 7.1 x 10[tex]^{3}[/tex]  (remove all digits after 7.1  

Hope this helps :)

Answer:

If b=5:7, find the value of B in terms of c

Answer:

you have to put the Y in the formula or make a comparison

Answer:

₹ 81.75

Step-by-step explanation:

Total =  ₹ 120.

He bought one book for ₹ 25.75 and a pen for ₹ 12.50.

    ₹ 25.75 + ₹ 12.50 =  ₹ 38.25

How much money does he have now?​

     ₹ 120 -  ₹ 38.25 =  ₹ 81.75

₹81.75

you just minus 120 from 25.75 and 12.50

Answer:

170°

Step-by-step explanation:

There is a total of 180° and the angle is 10° less...

So, 180°-10° = 170°

Please 5 stars if correct!!!

I hope this helps!!!

Answer: 3005

Step-by-step explanation: multiply 5 by 20 and it will give 100 then times 30 Equals 3000 and plus gives 3005 . Please mark brainliest

Answer:3005

Step-by-step explanation :multiply 5 by 20 and it will give 100 then multiply by 30  that Equals 3000 and plus 5 that gives you 3005

Answer:390

Step-by-step explanation:

500*.22=110

500-110=390

390

Answer:

390

Step-by-step explanation:

22% of 500 is 110 so that would make the answer 390 i think

A: 1.9 Kilograms of oranges for rs 665.

B: 12 liters of petrol will make a car travel 120 km.

option C

hope it helps...!!!

Answer:

[tex]\fbox{Area = 1017.36 sq.cm.}[/tex]

Step-by-step explanation:

Given:

circumference of circular dinner plate= 113.04 cm

To find:

Area of dinner plate = ?

Solution:

circumference = 113.04 cm.

we know that circumference = 2πr

[tex]2 \pi r = 113.04 \: cm \\ r = \frac{113.04 }{2 \pi} \\ r = \frac{113.04 }{2 \times 3.14} \\r = \frac{113.04 }{6.28} \\ \fbox{r = 18 \: cm}[/tex]

We know that,

Area of circle = πr²

now substituting value of r in above formula,

[tex] \pi {r}^{2} = 3.14 \times 18 \times 18 = 1017.36 \: {cm}^{2} [/tex]

[tex] \fbox{Area = 1017.36 sq.cm.}[/tex]

Thanks for joining brainly community!

Answer:

question 1 is 60 beads and question 2 is 16 dollars and 50 cents.

Step-by-step explanation:

for question 1 the are asking for the lowest common denomonator (60)

for question 2 you devide 60/20 and multiply by 5.50

3 x 5.50 = 16.5$

Answer: C

Step-by-step explanation:

Just look at a z-score table and multiply by 100.

-> (0.308538)(100) is about 30.85%

Answer:

ONE “1”

Step by step:

Taking into account the definition of a system of equations, this nonlinear system of equations has one solution.

A system of equations is a set of two or more equations that share two or more unknowns. The solutions of a system of equations are all the values that are valid for all equations.

That is, to find the solution to a system of equations, you must find a value (or range of values) that satisfies all the equations in the system.

Graphically, the points where the graphs of the equations intersect will be the solutions to the system.

A system of equations with a linear equation and a quadratic equation can have two, one, or no solutions. To find its quantity, we must look for the intersection between both graphs:

If the graphs of the equations do not intersect, then there are no solutions for both equations. If the parabola and the line touch at a single point, then there is a solution for both equations. If the line intersects the parabola in two places, then there are two solutions to both equations.

In this case, you can see that the parabola and the line touch at a single point. Then there is a solution for both equations.

In summary, this nonlinear system of equations has one solution.

Using the Pythagorean Theorem, it is found that Barrie is 12 miles from Luke's house, in a straight line.

The Pythagorean Theorem relates the length of the legs [tex]l_1[/tex] and [tex]l_2[/tex] of a right triangle with the length of the hypotenuse h, according to the following equation:

[tex]h^2 = l_1^2 + l_2^2[/tex]

This problem can be modeled by a right triangle, with [tex]l_1 = 16, l_2 = d, h = 20[/tex], hence:

[tex]h^2 = l_1^2 + l_2^2[/tex]

[tex]16^2 + d^2 = 20^2[/tex]

[tex]d^2 = 144[/tex]

[tex]d = 12[/tex]

Barrie is 12 miles from Luke's house, in a straight line.

More can be learned about the Pythagorean Theorem at https://brainly.com/question/654982

Answer:

i think the answer is d which is 50°

Answer:

[tex] \frac{ - 1}{6} [/tex]

Step-by-step explanation:

[tex]slope \: = \frac{y2 - y2}{x2 - x1} \\ slope = \frac{3 - 4}{5 - - 1} \\ sope = \frac{ - 1}{ \: \: \: 6} [/tex]

Answer:

2

Step-by-step explanation:

To find the interquartile range, subtract the lower quartile from the upper quartile.

The upper quartile is 7.

The lower quartile is 5.

Subtract.7–5=2

The interquartile range for Mike's teammates' shoe sizes is 2 sizes.

The interquartile range for Mike's teammates' shoe sizes is R = 2

The distance between the upper and lower quartiles is known as the interquartile range. Half of the interquartile range corresponds to the semi-interquartile range. Finding the values of the quartiles in a small data set is straightforward.

The spread of the data, or statistical dispersion, is measured by the interquartile range. The middle 50%, fourth spread, or H-spread are further names for the IQR. It is described as the spread between the data's 75th and 25th percentiles.

Given data ,

Let the interquartile range be represented as R

Now , the equation will be

Let the shoe sizes be represented as set A

A = { 2 , 4 , 6 , 8 , 10 }

Now , the upper range of the quartile is = 7

The lower range of the quartile is = 5

And , the interquartile range R = upper quartile - lower quartile

On simplifying , we get

The interquartile range R = 7 - 5 = 2

Hence , the interquartile range R = 2

To learn more about interquartile range click :

https://brainly.com/question/15686744

#SPJ3

Answer:

Yes

Step-by-step explanation:

An average of five birds hit the monument each week and in order to receive the funds, more than three birds have to be killed in a week

Answer:

Choice C - Input: 2 Output: 6

Answer:

C

Step-by-step explanation:

evaluate the output for the given input values using the rule

choice A

5 × 5 - 4 = 25 - 4 = 21 ≠ 5

choice B

3 × 5 - 4 = 15 - 4 = 11 ≠ 7

choice C

2 × 5 - 4 = 10 - 4 = 6 ← equals the output value

Answer:

[tex](9x-1)^2[/tex]

Step by step explanation:

[tex]81x^2-18x+1\\\\=81x^2 - 9x-9x+1\\\\=9x(9x-1)-(9x-1)\\\\=(9x-1)(9x-1)\\\\=(9x-1)^2[/tex]

[tex] \Large \bold{SOLUTION\ 1:} [/tex]

[tex] \small \begin{array}{l} \text{First, we need to check if the given differential} \\ \text{equation is exact.} \\ \\ (1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \quad M(x, y) dx + N(x, y) dy = 0 \end{array} [/tex]

[tex] \small \begin{array}{l l}\tt\: M(x,y) = \dfrac{1}{(1 - xy)^2}, & N(x,y) = y^2 + \dfrac{x^2}{(1-xy)^2}\\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{-2(-x)}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^2} + \dfrac{-2(-y)x^2}{(1 - xy)^3} \\ \\\tt \dfrac{\partial M}{\partial y} = \dfrac{2x}{(1 - xy)^3}, & \dfrac{\partial N}{\partial x} = \dfrac{2x(1 - xy)+2x^2y}{(1 - xy)^3} \\ \\\tt \: & \dfrac{\partial N}{\partial x} = \dfrac{2x}{(1 - xy)^3} \end{array} [/tex]

[tex] \small \begin{array}{l} \tt\dfrac{\partial M}{\partial y} = \dfrac{\partial N}{\partial x} \implies \text{Differential equation is exact.} \\ \\\tt \dfrac{\partial F}{\partial x} = M(x, y) = \dfrac{1}{(1 - xy)^2} \\ \tt\displaystyle F(x, y) = \int \dfrac{1}{(1 - xy)^2} \partial x = -\dfrac{1}{y} \int \dfrac{1}{(1 - xy)^2}(-y)\partial x \\ \\ \tt\:F(x, y) = \dfrac{1}{y(1 - xy)} + h(y) \\ \\ \tt\dfrac{\partial F}{\partial y} = N(x, y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\\tt \dfrac{\partial}{\partial y}\left[\dfrac{1}{y(1 - xy)} + h(y)\right] = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - xy + y(-x)}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ \tt-\dfrac{1 - 2xy}{y^2(1 - xy)^2} + h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} \\ \\ h'(y) = y^2 + \dfrac{x^2}{(1-xy)^2} + \dfrac{1 - 2xy}{y^2(1 - xy)^2} \\ \\ \tt\:h'(y) = y^2 + \dfrac{x^2y^2 - 2xy + 1}{y^2(1-xy)^2} = y^2 + \dfrac{1}{y^2} \\ \\ h(y) = \dfrac{y^3}{3} - \dfrac{1}{y} + C \\ \\ \tt\text{Substituting to }F(x,y),\text{we get} \\ \\ \dfrac{1}{y(1 - xy)} + \dfrac{y^3}{3} - \dfrac{1}{y} = C \\ \\ \quad \quad \text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

[tex] \Large \bold{SOLUTION\ 2:} [/tex]

[tex] \small \begin{array}{l} \tt\text{Since we already know that the equation is exact,} \\ \text{we can then continue solving for the solution by} \\ \text{inspection method or by algebraic manipulation.} \\ \\ \tt(1-xy)^{-2} dx + \big[y^2 + x^2 (1-xy)^{-2}\big]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + \left[y^2 + \dfrac{x^2}{(1-xy)^2}\right]dy = 0 \\ \\ \tt\dfrac{dx}{(1-xy)^2} + y^2 dy + \dfrac{x^2}{(1-xy)^2} dy = 0 \\ \\ \tt\dfrac{dx + x^2dy}{(1-xy)^2} + y^2 dy = 0 \\ \\ \tt\text{Divide both numerator and denominator of the} \\ \tt\text{fraction by }x^2. \end{array} [/tex]

[tex] \small \begin{array}{c}\tt \dfrac{\dfrac{1}{x^2}dx + dy}{\dfrac{(1-xy)^2}{x^2}} + y^2 dy = 0 \\ \tt\\ \tt\dfrac{\dfrac{1}{x^2}dx + dy}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt-\dfrac{\left(-\dfrac{1}{x^2}dx - dy\right)}{\left(\dfrac{1}{x}-y\right)^2} + y^2 dy = 0 \\ \\ \tt\displaystyle {\large{\int}} -\frac{d\left(\dfrac{1}{x}-y\right)}{\left(\dfrac{1}{x}-y\right)^2} + \int y^2 dy = \int 0 \\ \\ \tt\implies\tt \dfrac{1}{\dfrac{1}{x}-y} + \dfrac{y^3}{3} = C \\ \\\text{or} \\ \\ \tt\red{\boxed{\dfrac{x}{1 - xy} + \dfrac{y^3}{3} = C} \Longleftarrow \textit{Answer}} \end{array} [/tex]

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