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3 years ago

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Let (4.0, 8.32) represent an interval on the number line. Complete parts (a) and (b) below. (a) Find the value that is in

the middle of the interval. The value is nothing. (Type an integer or a decimal. Do not round.) (b) Find the distance from the middle of the interval to either endpoint. The distance is 4.32. (Type an integer or a decimal. Do not round.)

1 answer:

Molodets [167]3 years ago

7 0

Answer:a

Step-by-step explanation:

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How many solutions are there to the equation shown below?<br><br> 7x + 12 = 5x – 8

ExtremeBDS [4]

There is 1 solution. X = -10

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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.

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4 years ago

The area model represents the quotient of 108 ÷ 9. Which equation shown by the area model can be used to find the quotient?

Rzqust [24]

Solution :

Given that the area model that represents a quotient of 108 ÷ 9.

So, we multiply 9 by multiples of $10$ to $\text{estimate the quotient.}$

10 20 30 40

90 180 270 360

Because 108 is between 90 and 180, the quotient lies between 10 and 20.

Using 10 as the first partial quotient in the area model for 108 ÷ 9,

9 x 10 = 90 , so 108 - 90 = 18

Now, 9 x 2 = 18, so 18 - 18 = 0

∴ 10 + 2 = 12

Therefore, the quotient of 108 ÷ 9 is 12.

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3 years ago

Use the graph below to answer the following questions.

Maru [420]

Answer:

a) 3
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Step-by-step explanation:

Refer to attached for points

a) f(-1) = 3 as per graph

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We add horizontal line through point y=6 to find intersections with the graph

Take x-values of those points. The points are x = -3, x = -2, x = 2

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4 years ago

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lozanna [386]

Answer: 42.3 ft

<u>Step-by-step explanation:</u>

Draw a picture (see attachment)

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$\sin\theta=\dfrac{opposite}{hypotenuse}\quad \rightarrow \quad \sin25^o=\dfrac{y}{100}\quad \rightarrow \quad 100\ \sin 25^o=y\\$

$\large\boxed{42.3\ ft}=y$

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