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2 years ago

What common angle do ACE and ACD share?

Mathematics

2 answers:

SCORPION-xisa [38]2 years ago

6 0

The answer is the second option, <ACE = <ECA.

This is because C is in the middle of both, so it's referring to the same angle. The other options are all comparing different angles, that aren't congruent, so they're incorrect.

never [62]2 years ago

3 0

Answer:

∠ACE≅∠ECA

Step-by-step explanation:

If we take ΔACE and ΔACD,

∠ACE and ∠ECA are the common angle as compared to the rest one given.

If we take, ∠AFB and ∠EFD, both the angles are not contained in the given triangle.

Similarly, if we take ∠EBC and∠ECB, the ∠EBC is not included in the triangle ACE.

In, ∠CAD and ∠CEB, both the angles cannot be equal.

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The formula for the area of a triangle is a = bh, where b is the length of the base and h is the height. the equation solved for

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2 years ago

Write (5,0) with a slope of 2 in point-slope form.

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Y = mx + c
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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 $\vec b$ = $\frac{1}{5} \hat i+\frac{3}{5} \hat j$ .

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3 years ago

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Answer:

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2 years ago

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olga55 [171]

Step-by-step explanation:

9(x + 7)² divided by 3(x - 1)(x + 7)

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