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3 years ago

13

The sum of Jada and Sophie's ages is, at most, 18. Sophie is 3 years older than twice Jada's age. Write an inequality model Jada

's age.

1 answer:

liq [111]3 years ago

5 0

Answer:

Step-by-step explanation:

$s=2j+3\\ \\ j+s\le 18\\ \\ j+2j+3\le 18\\ \\ 3j+3\le 18\\ \\ 3j\le 15\\ \\ j\le 5\\ \\ \text{Though j=5 is the only solution to both equations.}$

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What is the median score achieved by a student who recorded the following scores on 10 math quizzes 68 55 70 62 71 58 81 82 63 7

Vinil7 [7]

Answer:

b. 68

Step-by-step explanation:

7 0

3 years ago

A)0.5<br> B)2.5<br> C)-0.5<br> D)-2.5

7nadin3 [17]

Answer:

B)2.5

Step-by-step explanation:

When x = 3 the actual = 6

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3 years ago

A) If a + 5 = 15, then find the value of a

ozzi

Answer:

The answer is 10

Step-by-step explanation:

a + 5 =15

15-5=10

6 0

3 years ago

Read 2 more answers

Which expression is equivalent??? help!

dexar [7]

Answer:

The equivalent expression for the given expression $\sqrt[3]{256x^{10}y^{7} }$ is

$4x^{3} y^{2}(\sqrt[3]{4xy} )$

Step-by-step explanation:

Given:

$\sqrt[3]{256x^{10}y^{7} }$

Solution:

We will see first what is Cube rooting.

$\sqrt[3]{x^{3}} = x$

Law of Indices

$(x^{a})^{b}=x^{a\times b}\\and\\x^{a}x^{b} = x^{a+b}$

Now, applying above property we get

$\sqrt[3]{256x^{10}y^{7} }=\sqrt[3]{(4^{3}\times 4\times (x^{3})^{3}\times x\times (y^{2})^{3}\times y )} \\\\\textrm{Cube Rooting we get}\\\sqrt[3]{256x^{10}y^{7} }= 4\times x^{3}\times y^{2}(\sqrt[3]{4xy}) \\\\\sqrt[3]{256x^{10}y^{7} }= 4x^{3}y^{2}(\sqrt[3]{4xy})$

∴ The equivalent expression for the given expression $\sqrt[3]{256x^{10}y^{7} }$ is

$4x^{3} y^{2}(\sqrt[3]{4xy} )$

5 0

3 years ago

Need the steps and ASAP

Mars2501 [29]

What’s the question theirs nothing

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3 years ago