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geniusboy [140]
3 years ago
11

What is the quotient of

Mathematics
1 answer:
PtichkaEL [24]3 years ago
8 0

Answer:

Step-by-step explanation:

1)-0.22

2) -0.055

3)0.61

4)-1.972

Can someone make sure for me? My calculators broken.

-viridiancat4, an 8th grader! :)

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-3z^7 - (-5z^7) = -3z^7 + 5z^7 = 2z^7
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2(log 18 - log 3) + = 2log 1/16
Semmy [17]
What do you need?

2log 3/8

= -0.85
4 0
2 years ago
The top and bottom margins of a poster are each 15 cm and the side margins are each 10 cm. If the area of printed material on th
12345 [234]

Answer:

the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

Step-by-step explanation:

From the given question.

Let p be the length of the of the printed material

Let q be the width of the of the printed material

Therefore pq = 2400 cm ²

q = \dfrac{2400 \ cm^2}{p}

To find the dimensions of the poster; we have:

the length of the poster to be p+30 and the width to be \dfrac{2400 \ cm^2}{p} + 20

The area of the printed material can now be:  A = (p+30)(\dfrac{2400 }{p} + 20)

=2400 +20 p +\dfrac{72000}{p}+600

Let differentiate with respect to p; we have

\dfrac{dA}{dp}= 20 - \dfrac{72000}{p^3}

Also;

\dfrac{d^2A}{dp^2}= \dfrac{144000}{p^3}

For the smallest area \dfrac{dA}{dp }=0

20 - \dfrac{72000}{p^2}=0

p^2 = \dfrac{72000}{20}

p² = 3600

p =√3600

p = 60

Since p = 60 ; replace p = 60 in the expression  q = \dfrac{2400 \ cm^2}{p}   to solve for q;

q = \dfrac{2400 \ cm^2}{p}

q = \dfrac{2400 \ cm^2}{60}

q = 40

Thus; the printed material has the length of 60 cm and the width of 40cm

the length of the poster = p+30 = 60 +30 = 90 cm

the width of the poster = \dfrac{2400 \ cm^2}{p} + 20 = \dfrac{2400 \ cm^2}{60} + 20  = 40 + 20 = 60

Hence; the dimension of the poster = 90 cm length and 60 cm  width i.e 90 cm by 60 cm.

4 0
3 years ago
An athlete knows that when she jogs along her neighborhood​ greenway, she can complete the route in 10 minutes. It takes 20 minu
IgorLugansk [536]

Answer:

10 miles per hour.

Step-by-step explanation:

Let x represent athlete's walking speed.

We have been given that her jogging rate is 5 mph faster than her walking​ rate, so athlete's jogging speed would be x+5 miles per hour.

\text{Distance}=\text{Rate}\cdot \text{Time}

10 minutes = 1/6 hour.

20 minutes = 1/3 hour

While walking, we will get D_{\text{walking}}=x\frac{\text{miles}}{\text{hour}}\cdot \frac{1}{3}\text{hour}

D_{\text{walking}}=\frac{x}{3}

While jogging, we will get D_{\text{jogging}}=(x+5)\frac{\text{miles}}{\text{hour}}\cdot \frac{1}{6}\text{hour}

D_{\text{jogging}}= \frac{(x+5)}{6}

Since athlete is covering same distance while walking and jogging, so we can equate both expressions as:

\frac{x}{3}=\frac{x+5}{6}

Cross multiply:

6x=3x+15

6x-3x=15

3x=15

\frac{3x}{3}=\frac{15}{3}\\\\x=5

Therefore, athlete's walking speed is 5 miles per hour.

Jogging speed: x+5\Rightarrow 5+5=10

Therefore, athlete's jogging speed is 10 miles per hour.

3 0
3 years ago
A $400 tv is on sale for 25% off what is the sale price of the tv
N76 [4]
300.

You can multiply by .75 to get the 25% off. 
5 0
3 years ago
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