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3 years ago

Please help me solve this problem

Mathematics

2 answers:

Rudiy273 years ago

8 0

It’s C (10 x 4) divided 2

cestrela7 [59]3 years ago

3 0

Answer:

C (10×4) ÷ 2

Step-by-step explanation:

C is the answer because if you do an equation,

10 multiplied by 4 (10×4)

Then, she wants to share the dolls equally for her brother

This means that you have to multi[ly 4 by 10, so you get 40. If she shares the dolls equally with her brother, she will give him 20 and keep 20 for herself.

C provides the correct answer with 20. 40/2 is 20

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"<br> Use the following graph to determine the solution(s) of:<br> y= x^2- x - 6

wariber [46]

Answer:

-2 and 3

Step-by-step explanation:

The x-intercepts or where the graph cuts the x-axis represents the zeros of y = x² - x - 6.

Thus, the zeros are -2 and 3. This is where the graph cuts the x-axis.

Therefore, the solutions of the graph are -2 and 3.

6 0

3 years ago

I GIVE BRAINLIEST FOR EXPLANATION AND CORRECT ANSWER EXTRA POINTS

madreJ [45]

Answer:x=4.2643

Step-by-step explanation:

tan35degrees=x/9

9(tan35degrees)=x

4.2643=x

6 0

3 years ago

Read 2 more answers

Please help math questions <br> put number with answers please

natita [175]

Answer:

See below for answers

Step-by-step explanation:

<u>Problem 1</u>

<u /> $\frac{1}{sin\theta}=2cos\theta\: ; \: 0\leq\theta < 2\pi\\\\1=2sin\theta cos\theta\\\\1=sin2\theta\\\\\frac{\pi}{2}+2\pi n=2\theta\\ \\\frac{\pi}{4}+\pi n=\theta\\ \\\theta=\bigr\{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\bigr\}$

Therefore, the second option is correct.

<u>Problem 2</u>

<u /> $cos(2x)+1=sin(x)+2\:;\: 0\leq x < 2\pi\\\\1-2sin^2x+1=sin(x)+2\\\\-2sin^2x=sin(x)\\\\0=2sin^2x+sinx\\\\0=sinx(2sinx+1)$

$0=sinx\\\\x=\pi n\\\\x=0,\pi$

$0=2sinx+1\\\\-1=2sinx\\\\-\frac{1}{2}=sinx\\ \\x=\frac{7\pi}{6}+2\pi n,\frac{11\pi}{6}+2\pi n\\\\x=\frac{7\pi}{6},\frac{11\pi}{6}$

Therefore, the solution set is $x=\bigr\{0,\pi,\frac{7\pi}{6},\frac{11\pi}{6}\bigr\}$ , making the second option correct.

<u>Problem 3</u>

<u /> $2cos^2x=-cosx\: ;\: 0^\circ\leq x < 360^\circ\\\\2cos^2x+cosx=0\\\\cosx(2cosx+1)=0$

$cosx=0\\\\x=\frac{\pi}{2}+\pi n\\ \\x=90^\circ+180n^\circ\\\\x=90^\circ,270^\circ$

$2cosx+1=0\\\\2cosx=-1\\\\cosx=-\frac{1}{2}\\\\x=\frac{2\pi}{3}+2\pi n,\frac{4\pi}{3}+2\pi n\\ \\x=120^\circ+360n^\circ,240^\circ+360n^\circ\\\\x=120^\circ,240^\circ$

Therefore, the solution set is $x=\bigr\{90^\circ,120^\circ,240^\circ,270^\circ\bigr\}$ , making the fourth option correct

7 0

2 years ago

V*w, vectors: r = <8, 1, -6>; v = <6, 7, -3>; w = <-7, 5, 2>

Misha Larkins [42]

<h2>Explanation:</h2><h2 />

Here we have to compute the dot product:

$v\cdot w$

We know that:

$v = \\ \\ w =$

So for any two vectors:

$A = \\ \\ B =$

The dot product:

$A\cdot B=x_{1}y_{1}+x_{2}y_{2}+z_{1}z_{2}$

Therefore:

$v\cdot w=(6)(-7)+(7)(5)+(-3)(2) \\ \\ \boxed{v\cdot w=-13}$

4 0

3 years ago

How would you draw a counterexample to this?

NeX [460]

Melanie said:
Every angle bisector in a triangle bisects the opposite side perpendicularly.

A 'counterexample' would show an angle bisector in a triangle that DOESN'T
bisect the opposite side perpendicularly.

See my attached drawing of a counterexample.

Both of the triangles that Melanie examined have equal sides on both sides
of the angle bisector. That's the only way that the angle bisector can bisect
the opposite side perpendicularly. Melanie didn't examine enough different
triangles.

6 0

4 years ago

Please help me solve this problem​

Please help me solve this problem