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3 years ago

17. Find the length of the midsegment XY.

Mathematics

1 answer:

fredd [130]3 years ago

4 0

Xy = 12 because if you look at the pattern the top one is 16 and the bottom one is 8 so the middle of those would be 12.

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The diagram shows squares arranged around the edge of a right triangle. What is the area of square B

Lina20 [59]

Step-by-step explanation:

take the right triangle

15^2 = 9^2 + x^2

225 = 81 + x^2

x = 12

area of square b = 144

6 0

2 years ago

Solve the equation by the Quadratic Formula. Round to the nearest tenth, if necessary. Write your solutions from least to greate

Marat540 [252]

I hope the choices for the numerators of the solutions are given.

I am showing the complete work to find the solutions of this equation , it will help you to find an answer of your question based on this solution.

The standard form of a quadratic equation is :

ax² + bx + c = 0

And the quadratic formula is:

x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

So, first step is to compare the given equation with the above equation to get the value of a, b and c.

So, a = 10, b = -19 and c = 6.

Next step is to plug in these values in the above formula. Therefore,

So,

Hope this helps you!

3 0

2 years ago

Solve for x in the equation 3 x squared minus 18 x + 5 = 47.

ryzh [129]

Answer:

x= 3 + $\sqrt{23}$ , 3 - $\sqrt{23}$

Step-by-step explanation:

You use the quadratic formula to get x= $\frac{18+6\sqrt{23} }{6} ,\frac{18-6\sqrt{23} }{6}$

Then you simplify and get the answers x= 3 + $\sqrt{23}$ , 3 - $\sqrt{23}$

6 0

3 years ago

Read 2 more answers

HELP FAST ALL OF MY POINTS TO BE GIVEN TO YOU

gogolik [260]

Answer:

C > 9

Step-by-step explanation:

3 0

3 years ago

What is the best approximation of the projection of (5,-1) onto (2,6)?

Hatshy [7]

Answer:

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

Step-by-step explanation:

We have given two points $(5, -1)$ and $(2, 6)$ .

Let, $\vec a=5\hat {i}-\hat {j}$ and $\vec b= 2\hat {i}+6\hat{j}$ .

and we have calculate the projection of $\vec a$ onto $\vec b$ .

Now,

For the calculation of projection, first we need to calculate the dot product of $\vec a$ and $\vec b$ .

$\vec a.\vec b=(5\hat {i}-\hat{j}).(2\hat{i}+6\hat{j})$

$=10-6$

$=4$

then, we have to calculate the magnitude of $\vec b$ .

$\mid {\vec {b}}\mid$ = $\sqrt{2^{2}+6^{2} }$ = $\sqrt{40}$ = $2\sqrt{10}$ .

Now, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\vec a.\vec b}{\mid b\mid}$

= $\frac{4}{2\sqrt{10} }$ $\frac{2}{\sqrt{10} } \times\frac{\sqrt{10} }{\sqrt{10} } =\frac{2\sqrt{10} }{10} = \frac{\sqrt{10} }{5}$

and the vector projection of $\vec a$ onto $\vec b$ = $\frac{\vec a. \vec b}{\mid\vec b \mid^{2} } . \vec b$

= $\frac{4}{40} . (2\hat i+ 6\hat j)$

= $\frac{1}{5} \hat i+\frac{3}{5} \hat j$

Hence, the scalar projection of $\vec a$ onto $\vec b$ = $\frac{\sqrt{10} }{5}$ , and the vector projection of $\vec a$ onto $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

6 0

3 years ago