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2 years ago

100 Points!!!!!!!!!!!! HELP

Mathematics

2 answers:

stira [4]2 years ago

8 0

I think it’s D?? But again I’m not sure sorry if you get it wrong tho

IrinaVladis [17]2 years ago

4 0

Answer:

d is the answer brainliest?

Step-by-step explanation:

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Solve the system by adding or subtracting.

irakobra [83]

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2 years ago

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Find and simplify the difference quotient of the form, <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bf%28x%2Bh%29-f%28x%29%7D%7Bh

Anastasy [175]

With $f(x)=-10x+9$ , we have

$f(x+h) = -10(x+h) + 9 = -10x + 9 - 10h$

Then the difference quotient is

$\dfrac{f(x+h)-f(x)}h = \dfrac{(-10x+9-10h)-(-10x+9)}h = \dfrac{-10h}h = -10$

since <em>h</em> ≠ 0.

7 0

2 years ago

Janiya just turned 18 years old. She spent 4/9 of her life in NY. How many years did Janiya live in NY? *

mr Goodwill [35]

4/9 x 2 = 8/18 so 8 yrs of her life

7 0

3 years ago

The angle of elevation from me to the top of a hill is 51 degrees. The angle of elevation from me to the top of a tree is 57 deg

julia-pushkina [17]

Answer:

Approximately $101\; \rm ft$ (assuming that the height of the base of the hill is the same as that of the observer.)

Step-by-step explanation:

Refer to the diagram attached.

Let $\rm O$ denote the observer.
Let $\rm A$ denote the top of the tree.
Let $\rm R$ denote the base of the tree.
Let $\rm B$ denote the point where line $\rm AR$ (a vertical line) and the horizontal line going through $\rm O$ meets. $\angle \rm B\hat{A}R = 90^\circ$ .

Angles:

Angle of elevation of the base of the tree as it appears to the observer: $\angle \rm B\hat{O}R = 51^\circ$ .
Angle of elevation of the top of the tree as it appears to the observer: $\angle \rm B\hat{O}A = 57^\circ$ .

Let the length of segment $\rm BR$ (vertical distance between the base of the tree and the base of the hill) be $x\; \rm ft$ .

The question is asking for the length of segment $\rm AB$ . Notice that the length of this segment is $\mathrm{AB} = (x + 20)\; \rm ft$ .

The length of segment $\rm OB$ could be represented in two ways:

In right triangle $\rm \triangle OBR$ as the side adjacent to $\angle \rm B\hat{O}R = 51^\circ$ .
In right triangle $\rm \triangle OBA$ as the side adjacent to $\angle \rm B\hat{O}A = 57^\circ$ .

For example, in right triangle $\rm \triangle OBR$ , the length of the side opposite to $\angle \rm B\hat{O}R = 51^\circ$ is segment $\rm BR$ . The length of that segment is $x\; \rm ft$ .

$\begin{aligned}\tan{\left(\angle\mathrm{B\hat{O}R}\right)} = \frac{\,\rm {BR}\,}{\,\rm {OB}\,} \; \genfrac{}{}{0em}{}{\leftarrow \text{opposite}}{\leftarrow \text{adjacent}}\end{aligned}$ .

Rearrange to find an expression for the length of $\rm OB$ (in $\rm ft$ ) in terms of $x$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{BR}}{\tan{\left(\angle\mathrm{B\hat{O}R}\right)}} \\ &= \frac{x}{\tan\left(51^\circ\right)}\approx 0.810\, x\end{aligned}$ .

Similarly, in right triangle $\rm \triangle OBA$ :

$\begin{aligned}\mathrm{OB} &= \frac{\mathrm{AB}}{\tan{\left(\angle\mathrm{B\hat{O}A}\right)}} \\ &= \frac{x + 20}{\tan\left(57^\circ\right)}\approx 0.649\, (x + 20)\end{aligned}$ .