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2 years ago

13

A map has a scale of 3 centimeters = 26 kilometers. If two cities are 7 centimeters apart on the map, what is the actual distanc

e between the two cities? If necessary, round your answer to the nearest tenth of a kilometer.

1 answer:

maw [93]2 years ago

5 0

Answer:

The answer to your question is 60.7 km

Step-by-step explanation:

This is a ratio and proportion problem, so let's write the proportion.

3 : 26 : : 7 : x

1.- Convert the expression to a fractional form.

$\frac{3}{26} = \frac{7}{x}$

2.- Solve the proportion, cross multiplying and back off.

x = $\frac{26 x 7}{3}$

x = $\frac{182}{3}$

x = 60.7 km

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Kristin spent $131 on shirts. Fancy shirts cost $28 and plain shirts cost $15. If she bought a total of 7 shirts then how many o

KIM [24]

Answer:

Kristin bought 5 plain shirts and 2 fancy shirts.

Hope this helps!

4 0

2 years ago

A baseball player has made 32 free throws in 45 attempts for current success rate of 71%. How many consecutive free throws must

s344n2d4d5 [400]

Answer: The player need to make 2 consecutive free throws to raise the average to 75%.

Step-by-step explanation:

Let the number of extra consecutive free throws to raise the average to 75% be 'x'.

Number of free throws already taken = 32

Number of attempts = 45

According to question, it becomes,

$\dfrac{32+x}{45}=0.75\\32+x=45\times 0.75\\\\32+x=33.75\\\\x=33.75-32\\\\x=1.75$

Hence, the player need to make 2 consecutive free throws to raise the average to 75%.

6 0

3 years ago

Select the two values of x that are roots of this equation x^2-5x+2=0

Nataly [62]

The roots of the equation is x = 4.56 OR x = 0.44

<h3>Quadratic equation</h3>

From the question, we are to determine the roots of the given equation

The given equation is

x² -5x +2 = 0

Using the formula method,

$x =\frac{-b \pm \sqrt{b^{2}-4ac } }{2a}$

In the given equation,

a = 1, b = -5, c = 2

Putting the values into the formula,

$x =\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(2) } }{2(1)}$

$x =\frac{5 \pm \sqrt{25-8} }{2}$

$x =\frac{5 \pm \sqrt{17} }{2}$

$x =\frac{5 + \sqrt{17} }{2}$ OR $x =\frac{5 - \sqrt{17} }{2}$

$x =\frac{5 + 4.12}{2}$ OR $x =\frac{5 - 4.12}{2}$

$x =\frac{9.12}{2}$ OR $x =\frac{0.88}{2}$

x = 4.56 OR x = 0.44

Hence, the roots of the equation is x = 4.56 OR x = 0.44

Learn more on Quadratic equation here: brainly.com/question/8649555

#SPJ1

3 0

2 years ago

What is the equation for p?

STALIN [3.7K]

F(x)=(x)-1 I think that’s the answer but use Desmos to figure out

6 0

2 years ago

B) Let g(x) =x/2sqrt(36-x^2)+18sin^-1(x/6)<br><br> Find g'(x) =

jolli1 [7]

I suppose you mean

$g(x) = \dfrac x{2\sqrt{36-x^2}} + 18\sin^{-1}\left(\dfrac x6\right)$

Differentiate one term at a time.

Rewrite the first term as

$\dfrac x{2\sqrt{36-x^2}} = \dfrac12 x(36-x^2)^{-1/2}$

Then the product rule says

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 x' (36-x^2)^{-1/2} + \dfrac12 x \left((36-x^2)^{-1/2}\right)'$

Then with the power and chain rules,

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12\left(-\dfrac12\right) x (36-x^2)^{-3/2}(36-x^2)' \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} - \dfrac14 x (36-x^2)^{-3/2} (-2x) \\\\ \left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-1/2} + \dfrac12 x^2 (36-x^2)^{-3/2}$

Simplify this a bit by factoring out $\frac12 (36-x^2)^{-3/2}$ :

$\left(\dfrac12 x(36-x^2)^{-1/2}\right)' = \dfrac12 (36-x^2)^{-3/2} \left((36-x^2) + x^2\right) = 18 (36-x^2)^{-3/2}$

For the second term, recall that

$\left(\sin^{-1}(x)\right)' = \dfrac1{\sqrt{1-x^2}}$

Then by the chain rule,

$\left(18\sin^{-1}\left(\dfrac x6\right)\right)' = 18 \left(\sin^{-1}\left(\dfrac x6\right)\right)' \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac x6\right)'}{\sqrt{1 - \left(\frac x6\right)^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18\left(\frac16\right)}{\sqrt{1 - \frac{x^2}{36}}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{3}{\frac16\sqrt{36 - x^2}} \\\\ \left(18\sin^{-1}\left(\dfrac x6\right)\right)' = \dfrac{18}{\sqrt{36 - x^2}} = 18 (36-x^2)^{-1/2}$

So we have

$g'(x) = 18 (36-x^2)^{-3/2} + 18 (36-x^2)^{-1/2}$

and we can simplify this by factoring out $18(36-x^2)^{-3/2}$ to end up with

$g'(x) = 18(36-x^2)^{-3/2} \left(1 + (36-x^2)\right) = \boxed{18 (36 - x^2)^{-3/2} (37-x^2)}$

5 0

2 years ago