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NemiM [27]
3 years ago
12

CAN SOMEONE PLEASE HELP ME WITH THIS PROBLEM​

Mathematics
1 answer:
padilas [110]3 years ago
7 0
I hope this helped
1 - i quadrant
2 - iii quadrant
3- y axis
4- iii quadrant
5- iv quadrant
6- x axis
7- i quadrant
8- ii quadrant
9- iv quadrant
10- ii quadrant
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Whitney works at an aquarium. She uses hoses of different sizes to fill the fish tanks. Hose A takes 16 min to fill a 160-gal ta
scZoUnD [109]

Answer:

Hose A fills at a rate of 10 gal per min

Hose B fills at a rate of 8 gal per min

Hose B fills at a faster rate

Step-by-step explanation:

7 0
2 years ago
Read 2 more answers
Which of the following best describes the solution to the equation below?
kozerog [31]
D)One solution. b=5
Distribute 2 times b plus 2 times 3
2b+6
2b+6+2b=26
Combine like terms 4b+6=26
Subtract 6 from both sides
4b=20
Divide both sides by 4
b=5

Check by substituting 5 for b
2 (5+3)+2(5)=26
2(8) + 2(5)=26
16+10=26
26=26
4 0
3 years ago
What is the value of the expression 2[4(2^3+5)]-4^2
ivolga24 [154]

Answer:

88

Step-by-step explanation:

Solve inside first

2^3=8

8+5= 13

Now multiply by 4

=52

Now by 2

=104

Now this is what we have left

=104-4^2

Make it easy by solving 4^2 which is 16.

104-16

=88

3 0
3 years ago
On a standardized exam, the scores are normally distributed with a mean of 300 and
Misha Larkins [42]

Answer:

-1.5

Step-by-step explanation:

8 0
3 years ago
Janeel has a 10-inch by 12-inch photograph. She wants to scan the photograph, then reduce the result
Vaselesa [24]

<u>Complete Question:</u>

Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.

<u>Correct Answer:</u>

A) (10-x)(12-x)=15

B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

<u>Step-by-step explanation:</u>

a. Write an equation to represent the area of the reduced image.

Let the reduced dimensions is by x , So the new dimensions are

length=10-x\\breadth=12-x

According to question , Area of new image is :

⇒ Area = \frac{1}{8}Length(breadth)

⇒ Area = \frac{1}{8}(10)(12)

⇒ Area = 15

So the equation will be :

⇒ (10-x)(12-x)=15

b. Find the dimensions of the reduced image

Let's solve :  (10-x)(12-x)=15

⇒ 120-10x-12x+x^2=15

⇒ 120-22x+x^2=15

⇒ x^2-22x+105=0

By Quadratic formula :

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

⇒ x = \frac{22 \pm8 }{2}

⇒ x = \frac{22 +8 }{2} , x = \frac{22 -8 }{2}

⇒ x = 15 , x =7

x = 15 is rejected ! as 15 > 10 ! Side can't be negative

⇒ x =7

Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

5 0
3 years ago
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