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kakasveta [241]
3 years ago
8

Mayumi plans to buy pencils and a notebook at the school store. A pencil costs $0.15, and a The notebook costs $1.59. Mayumi has

$5.00. Which inequality could she use to find the number of pencils she can buy? in equation please ​
Mathematics
1 answer:
Elina [12.6K]3 years ago
3 0

Answer: 0.15p+1.59n ≤ 5.00

Step-by-step explanation:

Given: A pencil costs $0.15, and a The notebook costs $1.59.

Let p = Number of pencils.

n = Number of notebooks.

Total cost of pencil and notebook = 0.15p+1.59n

Since Mayumi has $5.00.

So, Total cost of pencil and notebook ≤ $5.00

⇒ 0.15p+1.59n ≤ 5.00

Hence, the required inequality: 0.15p+1.59n ≤ 5.00

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James has 21 cups of flour. He will pour the flour equally into 4 containers. How many cups of flour will be in each container?
e-lub [12.9K]

Answer:

5.25 cups of flour

Step-by-step explanation:

4 containers - 21 cups of flour

1 container - 21 cups of flour ÷ 4 = 5.25 cups of flour

7 0
2 years ago
Read 2 more answers
Angle A and Angle B are vertical angles. If m
damaskus [11]

Answer:

∠B = 17°

Step-by-step explanation:

Vertical angles are congruent, thus

5x + 7 = x + 15 ( subtract x from both sides )

4x + 7 = 15 ( subtract 7 from both sides )

4x = 8 ( divide both sides by 4 )

x = 2

Hence

∠B = 5x + 7 = (5 × 2) + 7 = 10 + 2 = 17°

3 0
3 years ago
How are the values of -4^3 and 4^-3 alike and how do they differ?
BartSMP [9]

Answer:

(-4)³ = -64

(4)⁻³ = 1/64

Step-by-step explanation:

( - 4) {}^{3}  =  ( - 4) \times ( - 4) \times ( - 4) =  - 64 \\

{4}^{ - 3}  =  (\frac{1}{4} ) {}^{3}  \\

({ \frac{1}{4} )}^{3}  =  \frac{1}{4}  \times  \frac{1}{4}  \times  \frac{1}{4}  =  \frac{1}{64}  \\

4 0
2 years ago
4q=52 wats the answer to this equation ​
Darya [45]

4q=52

divide by 4 for both sides

4q/4= 52/4

q= 13

Answer : q=13

6 0
3 years ago
Note: Enter your answer and show all the steps that you use to solve this problem in the space provided.
prohojiy [21]

Answer:

2.25

Step-by-step explanation:

(6x^{-2})^2(0.5x)^4\\\\=(6^2(x^{-2})^{2})(0.5^4x^4)\ \ \ \ \ \ \ \ \ \ \ \ \ as\ (ab)^m=a^mb^m\\\\=36\times 0.5^4((x^{-2})^2x^4)\ \ \ \ \ \ \ \ \ \ as\ multiplication\ is\ associative\ a(bc)=(ab)c\\\\=36\times 0.0625(x^{-4}x^4)\ \ \ \ \ \ \ \ \ \ \ as\ (x^m)^n=x^{mn}\\\\=(36\times 0.0625)(x^{-4+4})\ \ \ \ \ \ \ \ \ \ \ as\ x^mx^n=x^{m+n}\\\\=2.25x^0\\\\=2.25\ \ \ \ \ \ \ \ \ \ \ \ as\ x^0=1\\\\(6x^{-2})^2(0.5x)^4=2.25

7 0
3 years ago
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