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3 years ago

Approximate Square root of 10 To the nearest 10th by hand to help you fill in the blanks

Mathematics

2 answers:

kaheart [24]3 years ago

8 0

The answer is 3.2! Hope this helps

geniusboy [140]3 years ago

4 0

Answer:

<h3>√10 = 3.2 approx. </h3>

Step-by-step explanation:

<h2>HOPE IT HELPS YOU!!</h2>

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What is the value of y when x = 3. y = 12x

alexandr1967 [171]

We have the equation y = 12x. And we know the value of x. The value of x is 3.
So, if we replace the x variable with the value of x, then we can solve the equation.

y = 12(3)
y = 36

So, the answer is 36.

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4 years ago

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Problem<br> 1<br> + 10x + 13 = 4

lutik1710 [3]

Answer:

$x = -1$

Step-by-step explanation:

Equation: $1+10x+13=4$

Step 1: Balance both parts of the equation.

$1+10x+13=4$

$(10x)+(1+13)=4$

$10x+14=4$

Step 2: Reduce 14 from both parts of the equation.

$10x+14-14=4-14$

$10x= -10$

Step 3: Split both sides by 10, because $\mathrn {10x = -10}$ .

$\frac{10x}{10} = \frac{-10}{10}$

$10 / -10 = -1$

Therefore, x = -1.

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3 years ago

Mr.morales makes a table of the number of chaperones needed for different number of students attending a field trip. the ratio o

vichka [17]

Answer:

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Step-by-step explanation:

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3 years ago

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What is the fourth term in the binomial expansion (a+b)^6)

Dafna11 [192]

Answer:

$20a^3b^3$

Step-by-step explanation:

<u>Binomial Series</u>

$(a+b)^n=a^n+\dfrac{n!}{1!(n-1)!}a^{n-1}b+\dfrac{n!}{2!(n-2)!}a^{n-2}b^2+...+\dfrac{n!}{r!(n-r)!}a^{n-r}b^r+...+b^n$

<u>Factorial</u> is denoted by an exclamation mark "!" placed after the number. It means to multiply all whole numbers from the given number down to 1.

Example: 4! = 4 × 3 × 2 × 1

Therefore, the fourth term in the binomial expansion (a + b)⁶ is:

$\implies \dfrac{n!}{3!(n-3)!}a^{n-3}b^3$

$\implies \dfrac{6!}{3!(6-3)!}a^{6-3}b^3$

$\implies \dfrac{6!}{3!3!}a^{3}b^3$

$\implies \left(\dfrac{6 \times 5 \times 4 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}{3 \times 2 \times 1 \times \diagup\!\!\!\!3 \times \diagup\!\!\!\!2 \times \diagup\!\!\!\!1}\right)a^{3}b^3$