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3 years ago

I have no clue how to do this

Mathematics

1 answer:

gregori [183]3 years ago

8 0

Me either just ask your teacher or use the app Socratic

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In a grade seven class, 18 students have pets, and 12 do not. What percent of the class has a pet?

Svetradugi [14.3K]

Answer :
If 18+12 is 30 and only 18 of them has pets then 70 percent have pets

7 0

2 years ago

Read 2 more answers

Spencer made a quilt for his cousin's doll. The quilt had a 7x7 array of different color square patches. If each patch is 1 3/4

Damm [24]

Answer:

$\frac{2401}{16}$ square inches

Step-by-step explanation:

Given:

The quilt made by Spencer had a 7×7 array of different color square patches such that each patch is $1\frac{3}{4}$ in long.

To find: the area of the whole quilt

Solution:

Side of each quilt = $7(\frac{3}{4})=7(\frac{7}{4})=\frac{49}{4}$

Quilt is in the form of a square such that area of square is given by $(side)^2$

Therefore, area of quilt = $(side)^2=(\frac{49}{4})^2 =\frac{2401}{16}$ square inches

3 0

3 years ago

A poster is to have a total area of 125 cm2. There is a margin round the edges of 6 cm at the top and 4 cm at the sides and bott

jeka57 [31]

Answer:

The correct answer is 15 cm.

Step-by-step explanation:

Let the width of the required poster be a cm.

We need to have a 6 cm margin at the top and a 4 cm margin at the bottom. Thus total margin combining top and bottom is 10 cm.

Similarly total margin combining both the sides is (4+4=) 8 cm.

So the required printing area of the poster is given by {( a-10 ) × ( a - 8) } $cm^{2}$

This area is equal to 125 $cm^{2}$ as per as the given problem.

∴ (a - 10) × (a - 8) = 125

⇒ $a^{2}$ - 18 a +80 -125 =0

⇒ $a^{2}$ - 18 a -45 = 0

⇒ (a-15) (a-3) = 0

By law of trichotomy the possible values of a are 15 and 3.

But a=3 is absurd as a $>$ 4.

Thus the required answer is 15 cm.

7 0

2 years ago

Read 2 more answers

I rlly need to know this asap

balu736 [363]

Box A has 12 balls and Box B has 8 balls.
OR
Box A has 60 balls and Box B has 40 balls.
OR
Box A has 48 balls and Box B has 32 balls.

7 0

2 years ago

If PN=29cm and MN=13 , then PM=?

HACTEHA [7]

*see attachment for the figure referred to

Answer/Step-by-step explanation:

1. PN = 29

MN = 13

PM = ?

$PN = PM + MN$ (Segment addition postulate)

$PN - MN = PM$ (subtract MN from each side)

$29 - 13 = PM$ (substitute)

$16 = PM$

$PM = 16 cm$

2. PN = 34, MN = 19, PM = ?

$PN = PM + MN$ (sediment addition postulate)

$PN - MN = PM$ (subtract MN from each side)

$34 - 19 = PM$ (substitute)

$15 = PM$

$PM = 15 cm$

3. PM = 19, MN = 23, PN = ?

$PN = PM + MN$ (Segment addition postulate)

$PN = 19 + 23$ (substitute)

$PN = 42 cm$

4. MN = 82, PN = 105, PM = ?

$PN = PM + MN$ (segment addition postulate)

$PN - MN = PM$ (subtract MN from each side)

$105 - 82 = PM$ (substitute)

$23 = PM$

$PM = 23 cm$

5. PM = 100, MN = 100, PN = ?

$PN = PM + MN$ (Segment addition postulate)

$PN = 100 + 100$ (substitute)

$PN = 200 cm$

7 0

2 years ago