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4 years ago

PLZ HELP I WILL GIVE BRAINLIEST..

Mathematics

1 answer:

cestrela7 [59]4 years ago

7 0

The answers are:

-9 ⁰ F
+9 pounds

I hope it helps

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Solve for x.<br> 10<br> 8<br> 6

Serga [27]

Answer:

Step-by-step explanation:

Intersecting Secants Theorem:

6(x + 6) = 8(8 + 10)

6x + 36 = 8(18)

6x + 36 = 144

6x = 144 - 36

6x = 108

x = 18

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3 years ago

Which is the graph of f（x）？

Kay [80]

Answer:

Third graph is the correct choice. See below.

Step-by-step explanation:

See the attachment below.

Best Regards!

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3 years ago

HELP I NEEEEEEEEEEEEEEEED HELP

Oksanka [162]

Answer:

It's B

Step-by-step explanation:

8 0

3 years ago

10. What is the area of a triangle with base 4cm and

sergiy2304 [10]

Step-by-step explanation:

Area of triangle = 0.5(4)(8) = 16cm² (D).

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3 years ago

HELP PICS INCLUDED! WILL GIVE BRAINLIEST! SHOW WORK AND EXPLAIN

Naddik [55]

I assume you know about the dot product, and that for two vectors $\mathbf a$ and $\mathbf b$ , the angle between them $\theta$ satisfies

$\mathbf a\cdot\mathbf b=\|\mathbf a\|\|\mathbf b\|\cos\theta\iff\cos\theta=\dfrac{\mathbf a\cdot\mathbf b}{\|\mathbf a\|\|\mathbf b\|}$

Then the vectors are parallel if the angle between them is 0 or 180 degrees (0 or pi radians), which would make $\cos\theta=1$ or $\cos\theta=-1$ , respectively.

Part A)

$\vec v_1=\langle\sqrt3,1\rangle\implies\|\vec v_1\|=\sqrt{(\sqrt3)^2+1^2}=\sqrt4=2$

$\vec v_2=\langle-\sqrt3,-1\rangle=-\vec v_1\implies\|\vec v_2\|=\|\vec v_1\|=2$

$\vec v_1\cdot\vec v_2=(\sqrt3)(-\sqrt3)+(1)(-1)=-4$

Then the angle between $\vec v_1,\vec v_2$ is such that

$\cos\theta=\dfrac{-4}{(2)(2)}=-1\implies\theta=\pi\,\mathrm{rad}$

so these vectors are parallel ("antiparallel", more specifically, which means they are parallel but point in opposite directions).

Part B) involves the same computations:

$\vec u_1=\langle2,3\rangle\implies\|\vec u_1\|=\sqrt{2^2+3^2}=\sqrt{13}$

$\vec u_2$ has the same components but differing by sign and order, as $\vec u_1$ ; its magnitude remains the same, though:

$\vec u_2=\langle-3,-2\rangle\implies\|\vec u_2\|=\sqrt{(-3)^2+(-2)^2}=\sqrt{13}$

$\vec u_1\cdot\vec u_2=(2)(-3)+(3)(-2)=-12$

$\implies\cos\theta=\dfrac{-12}{(\sqrt{13})(\sqrt{13})}=-\dfrac{12}{13}\implies\theta=\cos^{-1}\left(-\dfrac{12}{13}\right)$

which is neither 0 nor pi, which means these vectors are not parallel.

4 0

4 years ago