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3 years ago

14

Is this 117?? PLEASE HELP

1 answer:

natima [27]3 years ago

3 0

X is not 117 because 1 triangle is 180 degrees
So as you see in the picture x is a right triangle so it is 90 degrees

You might be interested in

The mass of an object is equal to the product of the object's density and volume. The density of titanium is 4.51 grams per cubi

natta225 [31]

284.13 grams

because you do 4.51 x 21 , then you have to multiply by 3

6 0

3 years ago

Could you please help me with this? It’s a problem from my grandson’s homework.

gulaghasi [49]

SOLUTION

From the question given, the cost of the toy truck is $36.50 before tax.

Now, the store selling the toy truck adds a 6% sales tax. This means

$\begin{gathered} 6\%\times36.50 \\ =\frac{6}{100}\times36.50 \\ =\frac{6\times36.50}{100} \\ =\frac{219}{100} \\ =2.19\text{ dollars} \end{gathered}$

Now, we will add this $2.19 to the initial cost, we have

$2.19+36.50=38.69$

This gives $38.69.

Finally we will do a discount of 10%. 10% discount is

$\begin{gathered} 10\%\times38.69 \\ \frac{10}{100}\times38.69 \\ =\frac{10\times38.69}{100} \\ =\frac{386.9}{100} \\ =3.869 \end{gathered}$

We will then subtract this 3.869 from 38.69, we get

$\begin{gathered} 38.69-3.869 \\ =34.821 \end{gathered}$

That is 34.821. The 1 which is the last number at the decimal part is less than 5, so we ignore it and our answer becomes $34.82

3 0

1 year ago

algol [13]

4/5 of 40 = 32
1/3 of 3 = 1
1/3 of 15 muffins = 10 muffins

3 0

3 years ago

**!!PLEASE ANSWER!!**

sammy [17]

Answer:

Answer: 155.12km

Step-by-step explanation:

sin(64˚) x 78 = 70.11 (2DP)

70.12 (2DP) + 85 = 155.1059356 (155.12 - 2DP)

5 0

3 years ago

Let f(x) = 1/x^2 (a) Use the definition of the derivatve to find f'(x). (b) Find the equation of the tangent line at x=2

Verdich [7]

Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

$f(x)=\frac{1}{x^2}$ .... (1)

According to the first principle of the derivative,

$f'(x)=lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{1}{(x+h)^2}-\frac{1}{x^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{\frac{x^2-(x+h)^2}{x^2(x+h)^2}}{h}$

$f'(x)=lim_{h\rightarrow 0}\frac{x^2-x^2-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-2xh-h^2}{hx^2(x+h)^2}$

$f'(x)=lim_{h\rightarrow 0}\frac{-h(2x+h)}{hx^2(x+h)^2}$

Cancel out common factors.

$f'(x)=lim_{h\rightarrow 0}\frac{-(2x+h)}{x^2(x+h)^2}$

By applying limit, we get

$f'(x)=\frac{-(2x+0)}{x^2(x+0)^2}$

$f'(x)=\frac{-2x)}{x^4}$

$f'(x)=\frac{-2)}{x^3}$ .... (2)

Therefore $f'(x)=-\frac{2}{x^3}$ .

(b)

Put x=2, to find the y-coordinate of point of tangency.

$f(x)=\frac{1}{2^2}=\frac{1}{4}=0.25$

The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

$y-y_1=m(x-x_1)$

$y-0.25=-0.25(x-2)$

$y-0.25=-0.25x+0.5$

$y=-0.25x+0.5+0.25$

$y=-0.25x+0.75$

Therefore the equation of the tangent line at x=2 is y=-0.25x+0.75.

5 0

3 years ago