Solve. 58 = y + 39 Enter your answer in the box. y =

Mathematics

1 answer:

Fudgin [204]2 years ago

6 0

58=y=39

(subtract 39 from both sides)

58-39=y

(simplify 58-39 to 19)

19=y

(switch sides)

y=19

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Susie has a bag with 8 hair pins, 7 pencils, 2 snacks, and 4 books. What is the ratio of books to pencils?

Viefleur [7K]

<h3>Hello there!</h3>

The first thing we need to do here is see which information is relevant.

The question is asking for the ratio of books to pencils.

Therefore, we only need the number of books and pencils - all other information given is irrelevant in this scenario.

Number of books - 4

Number of pencils - 7

Now, we can order this as a ratio.

A ratio is two numbers put as a proportion. It's normally written out as first number : second number.

In this case, it's the ratio of number of books : number of pencils.

Fill in the number of books and number of pencils into each side of the equation.

number of books : number of pencils

4 books : 7 pencils (the unit is normally dropped)

So therefore, 4:7 would be your final answer.

Hope this helped!

4 0

3 years ago

Read 2 more answers

At 2:00 PM a car's speedometer reads 30 mi/h. At 2:20 PM it reads 50 mi/h. Show that at some time between 2:00 and 2:20 the acce

Bad White [126]

Answer:

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$ . By the Mean Value Theorem, there is a number c such that $0 < c$ with $v'(c)=60 \:{\frac{mi}{h^2}}$ . Since v'(t) is the acceleration at time t, the acceleration c hours after 2:00 PM is exactly $60 \:{\frac{mi}{h^2}}$ .

Step-by-step explanation:

The Mean Value Theorem says,

Let be a function that satisfies the following hypotheses:

f is continuous on the closed interval [a, b].
f is differentiable on the open interval (a, b).

Then there is a number c in (a, b) such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Note that the Mean Value Theorem doesn’t tell us what c is. It only tells us that there is at least one number c that will satisfy the conclusion of the theorem.

By assumption, the car’s speed is continuous and differentiable everywhere. This means we can apply the Mean Value Theorem.

Let v(t) be the velocity of the car t hours after 2:00 PM. Then $v(0 \:h) = 30 \:{\frac{mi}{h} }$ and $v( \frac{1}{3} \:h) = 50 \:{\frac{mi}{h} }$ (note that 20 minutes is $20/60=1/3$ of an hour), so the average rate of change of v on the interval $[0 \:h, \frac{1}{3} \:h]$ is

$\frac{v(1/3)-v(0)}{1/3-0}=\frac{50 \:{\frac{mi}{h} }-30\:{\frac{mi}{h} }}{1/3\:h-0\:h} = 60 \:{\frac{mi}{h^2} }$

We know that acceleration is the derivative of speed. So, by the Mean Value Theorem, there is a time c in $(0 \:h, \frac{1}{3} \:h)$ at which $v'(c)=60 \:{\frac{mi}{h^2}}$ .