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2 years ago

5

Charlie has 324 magazines to pack in 5 boxes . He will pack the same number of magazines in each box. How many magazines will he

have left over

1 answer:

PtichkaEL [24]2 years ago

5 0

Answer:

4 magazines left over

Step-by-step explanation:

$\frac{324}{5} = 64 \frac{4}{5}$

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Kimora wants to put ribbon around her SQUARE picture frame. One side is 7 inches long. How much ribbon will she need? (hint: aro

denis23 [38]

Answer is 28 inches.

A square has equal sides and to find the perimeter, you add all the sides.

7 x 4 = 28

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2 years ago

Here is Triangle A. Lin created a scaled copy of Triangle A with an area of 72 square units. a. How many times larger is the are

PSYCHO15rus [73]

Answer:

The answer is below

Step-by-step explanation:

a) Triangle A is attached in the image below.

The base of triangle A is 3 units and its height is 3 units. The area of a triangle is given as:

Area = (1/2) × base × height

Area of triangle A = (1/2) × base × height = (1/2) × 3 × 3 = 4.5 unit²

Area of the scaled copy = 72 unit²

Ratio of area = Area of the scaled copy / Area of triangle A = 72 unit² / 4.5 unit² = 16

Hence the scaled copy area is 16 times larger than that of triangle A.

b) For the scaled copy:

Area of the scaled copy = (1/2) × base × height = 72 unit²

base × height = 144

Since the base and height are equal

base² = 144

base = 12, also height = 12

Base of scaled copy = 12 = 4 × base of triangle A

Therefore the scale factor used is 4

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2 years ago

What is each measure of the rhombus?

Elodia [21]

A. 15
b. 26

d. 41
e. 82
f. 98

h. 49

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2 years ago

in a AP the first term is 8,nth term is 33 and sum to first n terms is 123.Find n and common difference

allsm [11]

I believe there is no such AP...

Recursively, this sequence is supposed to be given by

$\begin{cases}a_1=8\\a_k=a_{k-1}+d&\text{for }k>1\end{cases}$

so that

$a_k=a_{k-1}+d=a_{k-2}+2d=\cdots=a_1+(k-1)d$

$a_n=a_1+(n-1)d$

$33=8+(n-1)d$

$21=(n-1)d$

$n$ has to be an integer, which means there are 4 possible cases.

Case 1: $n-1=1$ and $d=21$ . But

$\displaystyle\sum_{k=1}^2(8+21(k-1))=37\neq123$

Case 2: $n-1=21$ and $d=1$ . But

$\displaystyle\sum_{k=1}^{22}(8+1(k-1))=407\neq123$

Case 3: $n-1=3$ and $d=7$ . But

$\displaystyle\sum_{k=1}^4(8+7(k-1))=74\neq123$

Case 4: $n-1=7$ and $d=3$ . But

$\displaystyle\sum_{k=1}^8(8+3(k-1))=148\neq123$

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2 years ago

Solve for x :<br><br>14x + 2 = 9x - 3

exis [7]

The answer for your question is x=-1

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1 year ago