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2 years ago

SOMEONE PLEASE PLEASE HELP ME ON THIS!!

Mathematics

2 answers:

erik [133]2 years ago

5 0

$\frac{x}{20} =\frac{14}{8} \\x=\frac{14}{8} \times 20=35$ Answer:

Step-by-step explanation:

MArishka [77]2 years ago

4 0

X=22

14 plus 8 gives you x

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There are 9 tennis balls in 3 cans. What is the rate of number of tennis balls to number of cans?

garri49 [273]

ivied the total number of tennis balls buy the number of cans to see how many tennis balls are in 1 can

9 divided by 3 = 3

there are 3 tennis balls per can

fraction would be 3/1 ( meaning 3 balls to 1 can)

6 0

3 years ago

Please help <br> I will mark brainlist

Naya [18.7K]

Answer:

B x = sqrt(48)

Step-by-step explanation:

Half of the bottom side measures 4.

4^2 + x^2 = 8^2

16 + x^2 = 64

x^2 = 48

x = sqrt(48)

Answer: B x = sqrt(48)

4 0

3 years ago

Help please ! solve for x and round the answer to the nearest tenth. need it asap , thank you

mafiozo [28]

Answer:

3.6

Step-by-step explanation:

11x=40

X=40/11

6 0

2 years ago

Jamie takes at random a counter from the bag and records the colour. Red : blue : green = 1:3:7 Jamie does this a number of time

Sonbull [250]

Answer: There are 249 times Jamie takes a counter from the bag.

Explanation:

Since we have given that

Ratio of colour of red to blue to green is 1:3:7.

Number of blue counters = 68

Let the total number of times Jamie does from the bag be x.

So, According to question,

$\frac{3}{11}\times x=68\\\\x=\frac{11}{3}\times 68\\\\x=249.33\\\\=249\ approx.$

Hence, there are 249 times Jamie takes a counter from the bag.

8 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