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3 years ago

Simplify the expression by combining like terms :

1 answer:

8 0

Here is your answer:

2b + 7b^2 - 2 + 3b - 1 + b
Combine 2b and 3b
2b+3b=5b
Then do the exponent rule:
7×7=49b
Which equals:

(5b) b^2 + ( 2 ) b + (4b)

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If the temperature drops from 7f to -17f how much did the temperature decrease

Ipatiy [6.2K]

Answer:

there was a decrease of 24 degrees.

Step-by-step explanation:

7-24=-17

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3 years ago

Read 2 more answers

What is the slope of the equation Y=5/4x-7/4?

maksim [4K]

Answer:

Step-by-step explanation:

Compare the equation with y = mx + b where m is slope and b is y intercept

y = (5/4)x - (7/4)

Slope = m = 5/4

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2 years ago

Debra borrowed $8000 at a rate of 18% compounded semiannually. Assuming she makes no payment, how much will she owe after 9 year

Marina CMI [18]

Answer:

She owe $37736.96 after 9 years .

Step-by-step explanation:

Debra borrowed $8000 at a rate of 18% compounded semiannually

We are supposed to find how much will she owe after 9 years

Principal = 8000

Rate of interest = 18% =0.18

No. of compounds per year = 2

Time = 9 years

Formula : $A= P(1+\frac{r}{n})^{nt}$

Substitute the values in the formula :

$A= 8000(1+\frac{0.18}{2})^{2 \times 9}$

A= 37736.96

Hence She owe $37736.96 after 9 years .

5 0

3 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.