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3 years ago

I need help with this one

Mathematics

1 answer:

nikklg [1K]3 years ago

8 0

x-intercept when y = 0

Look at the graph, when y = 0, x = 2

So x-intercept (2, 0)

Answer

C. (2,0)

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What is the value of x in the circle on the right

german

To solve this we need to use Pythagorean theorem

${x}^{2} + {14}^{2} = (9 + x) ^{2} \\ {x}^{2} + 196 = 81 + 18x + {x}^{2} \\ 18x = 115 \\ x = 6.3(8) \simeq6.4$

Answer: x≈6.4

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3 years ago

Need to know the intercepts for x and y

Nataly [62]

X-intercept: -1
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2 years ago

I will give you BRAINLIEST for the correct answer

Verizon [17]

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3 years ago

What is the answer for 1+1+7x2+7y3+3+2x2−y3

natta225 [31]

First, a little housekeeping!!

1+1+7x2+7y3+3+2x2−y3 needs to be written as <span>1+1+7x^2+7y^3+3+2x^2−y^3. The " ^ " signifies exponentiation and is essential here.

Now combine like terms:

7y^3-y^3 + 7x^2 + 2x^2 + 1 + 1 + 3

You'll get 6y^3 +9x^2 + 5.</span>

6 0

3 years ago

The shadow of a tower at a time is three times as long as its shadow when the angle of elevation of the Sun is 60°. Find the ang

Naddika [18.5K]

Answer:

$30^{\circ}$ .

Step-by-step explanation:

Let $\theta$ denote the unknown angle of elevation. Let $h$ denote the height of the tower.

Refer to the diagram attached. In this diagram, ${\sf A}$ denotes the top of the tower while ${\sf B}$ denote the base of the tower; ${\sf BC}$ and ${\sf BD}$ denote the shadows of the tower when the angle of elevation of the sun is $60^{\circ}$ and $\theta$ , respectively. The length of segment ${\sf AB}$ is $h$ ; $\angle {\sf ACB} = 60^{\circ}$ , $\angle {\sf ADB} = \theta$ , and ${\sf BD} = 3\, {\sf BC}$ ..

Note that in right triangle $\triangle {\sf ABC}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ACB}$ . At the same time, segment ${\sf BC}$ (shadow of the tower when the angle of elevation of the sun is $60^{\circ}$ ) is adjacent to $\angle {\sf ACB}$ .

Thus, the ratio between the length of these two segments could be described with the tangent of $m\angle {\sf ACB} = 60^{\circ}$ :

$\begin{aligned}\tan(\angle {\sf ACB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AB}}{{\sf BC}}\end{aligned}$ .

$\begin{aligned}\frac{{\sf AB}}{{\sf BC}} = \tan(60^{\circ}) = \sqrt{3}\end{aligned}$ .

The length of segment ${\sf AB}$ is $h$ . Rearrange this equation to find the length of segment ${\sf BC}$ :

$\begin{aligned} {\sf BC} &= \frac{{\sf AB}}{\tan(\angle ACB)} \\ &= \frac{h}{\tan(60^{\circ})}\\ &= \frac{h}{\sqrt{3}} \\ &\end{aligned}$ .

Therefore:

$\begin{aligned}{\sf BD} &= 3\, {\sf BC} \\ &= \frac{3\, h}{\sqrt{3}} \\ &= (\sqrt{3})\, h\end{aligned}$ .

Similarly, in right triangle ${\sf ABD}$ , segment ${\sf AB}$ (the tower) is opposite to $\angle {\sf ADB}$ . Segment ${\sf BD}$ (shadow of the tower, with $\theta$ as the angle of elevation of the sun) is adjacent to $\angle {\sf ADB}$ .

$\begin{aligned}\tan(\angle {\sf ADB}) &= \frac{\text{opposite}}{\text{adjacent}} = \frac{{\sf AD}}{{\sf BD}}\end{aligned}$ .

$\begin{aligned}\frac{{\sf AB}}{{\sf BD}} = \tan(\theta) \end{aligned}$ .

Since ${\sf AB} = h$ while ${\sf BD} = (\sqrt{3})\, h$ :

$\begin{aligned}\tan(\theta) &= \frac{{\sf AB}}{{\sf BD}} \\ &= \frac{h}{(\sqrt{3})\, h} \\ &= \frac{1}{\sqrt{3}}\end{aligned}$ .

Therefore:

$\begin{aligned}\theta &= \arctan\left(\frac{1}{\sqrt{3}}\right) \\ &= 30^{\circ}\end{aligned}$ .

In other words, the angle of elevation of the sun at the time of the longer shadow would be $30^{\circ}$ .

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2 years ago