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Gnesinka [82]
4 years ago
12

Help on both plz!!!!!!

Mathematics
1 answer:
lilavasa [31]4 years ago
3 0
As long as the picture is in a good area where some one can make an educated guess on what text goes with it, I'd say have the text go below the picture
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Rachel and Hugo sorted 236 crayons into boxes for a local arts project. Each box had 10 crayons. How many crayons were left over
yarga [219]
230/10 = 23 with no remainder.
236/10 = 23 with remainder 6.
6 crayons were left over.
5 0
3 years ago
The height of a soccer ball that is kicked from the ground can be approximated by the function:
AfilCa [17]

Answer:

\displaystyle \sqrt{3}\:sec.

Step-by-step explanation:

\displaystyle -54 = -18x^2 → \frac{-54}{-18} = \frac{-18x^2}{-18} \\ \\ 3 = x^2 → \sqrt{3} = x

I am joyous to assist you anytime.

6 0
3 years ago
Solve x - 6 = -12 A) -18 B) -6 C) 2 D) 6
asambeis [7]
The answer would be -6 or letter B because
x - 6 =   - 12 \\ x - 6 + 6 =  - 12 + 6 \\ x =  - 6
8 0
3 years ago
Read 2 more answers
In Lily's garden, there are 5 rose bushes the first year. Each year, she adds two new rose bushes. She
aniked [119]

Step-by-step explanation:

Let x = the number of years since the first year and let y = the total number of plants.

Roses: y = 5 + 2x

Tulips: y = 20 – 3x

You can use elimination to solve.

y = 5 + 2x

(-)y = 20 – 3x

0 = -15 + 5x

15 = 5x

3 = x

This means that 3 years after the first year, the number of rose bushes equals the number of tulip plants.

The graphs appear to

intersect at (3, 11) which verifies that x=3

7 0
3 years ago
Read 2 more answers
David found and factored out the GCF of the polynomial 80b4 – 32b2c3 + 48b4c. His work is below.
Vikentia [17]

Answer:

Given Polynomial:

80b^4-32b^2c^3+48b^4c

Factors of Coefficient of terms

80 = 5 × 16

32 = 2 × 16

48 = 3 × 16

Common factor of the coefficient of all term is 16.

Each term contain variable. So the Minimum power of b is common from all terms.

Common from all variable part comes b².

So, Common factor of the polynomial = 16b²

⇒ 16b² ( 5b² ) - 16b² ( 2c³ ) + 16b² ( 3b²c )

⇒ 16b² ( 5b² - 2c³ + 3b²c )

Therefore, Statements that are true about David's word are:

The GCF of the coefficients is correct.

The variable c is not common to all terms, so a power of c should not have been factored out.

In step 6, David applied the distributive property

8 0
3 years ago
Read 2 more answers
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