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Dima020 [189]
4 years ago
9

Find the product. (3a^4 + 4)^2

Mathematics
2 answers:
Ira Lisetskai [31]4 years ago
6 0
9a8+24a4+16

Make sure that 8 and 4 are to the power
kirill115 [55]4 years ago
5 0

Answer: The product is 9a^8+24a^4+16

Step-by-step explanation:

Since we have given that

(3a^4+4)^2

As we know that

(a+b)^2=a^2+2ab+b^2

So, it becomes,

(3a^4 + 4)^2=(3a^4)^2+2\times 3a^4\times 4+(4)^2\\\\=9a^8+24a^4+16

Hence, the product is 9a^8+24a^4+16

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Please solve and explain<br><br> 1/5x = 20<br> x = ?
elixir [45]

Answer:

x = 1/100

Step-by-step explanation:

see the pic for the steps

5 0
4 years ago
im rlly confused (its yes and no answers)Decide if each pair of expressions is equivalent.6(-4t + 5) and -2(15 - 12t)1/5(20t - 1
kirill [66]

Given data:

The expression of the first option is,

\begin{gathered} \text{6(-4t+5})and\text{-2(15}-12t) \\ =-6(5-4t) \end{gathered}

Thus, the above expression are not equivalent.

The expression in the second option is,

\begin{gathered} \frac{1}{5}(20t-15)\text{ and }4t-3 \\ 4t-3\text{ and 4t-3} \end{gathered}

Thus, yes they are equivalent.

The expression in the third option is,

\begin{gathered} 1.5\text{ }-2\text{t and 0}.25(8-6t) \\ 1.5\text{ }-2\text{t and 2-1.5t} \end{gathered}

Thus, the above expression are not equivalent.

The expression in fourth option is,

\begin{gathered} -3(t-7+5t+8)\text{ and }-18t-3 \\ -3(6t+1)\text{ and -18t-3} \\ -18t-3\text{ and -18t-3} \end{gathered}

Thus, yes the above expression is equivalent.

3 0
2 years ago
Monica’s school band held a car wash to raise money for a trip to a parade in New York City. After washing 125 cars, they made $
mafiozo [28]

Answer:

5x + 8y = 775 and x + y =125

Step-by-step explanation:

3 0
3 years ago
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2<br> y + 7 = 5(x - 3)<br> What is the standard form for this?
lesya [120]

Answer:

y=10/3 i think if that's what youre asking

Step-by-step explanation:

4 0
3 years ago
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Write an expression that, when simplified, is equivalent to 15x + 7
Bas_tet [7]
15 times x plus 7
I hope i helped!
8 0
4 years ago
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