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2 years ago

6

Why did the football coach send in his second string

1 answer:

lorasvet [3.4K]2 years ago

3 0

Because his first string was tied up at the moment

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A package of hamburgers contains 8 patties and costs $7.50. Part A.) Luna has to buy at least 16 packages for an upcoming picnic

Rus_ich [418]

Answer:

Part A) $p\geq 128\ patties$

Part B) She will spend more than $\$142.50$

Step-by-step explanation:

Let

p-----> the number of hamburger patties

Part A) Luna has to buy at least 16 packages for an upcoming picnic

$p\geq 16*8$

$p\geq 128\ patties$

Part B) Suppose she actually needs more than 150 hamburgers. How much will she spend?

Let

c---------> the total cost

step 1

Divide 150 hamburgers by 8 (a package of hamburgers)

so

$\frac{150}{8} =18.75\ package$

round to the nearest whole number

$18.75=19\ package$ ----> the minimum number of packages

step 2

$c>19*\$7.50$

$c>\$142.50$

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2 years ago

Ronald spins a spinner that has 10 sections numbered from 1 to 10. What is the probability that the spinner lands on a prime num

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The probability the spinner lands on a prime number is 4/10 or 40%

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2 years ago

Read 2 more answers

Find the square root of 15129 by division method

MAXImum [283]

$\underline{\ \ \ \ \ \ 123}\\1\ \ \ |15129\\\underline{\ \ \ \ \|1}\\22\ |\ 51\\\underline{\ \ \ \ |\ 44}\\243|\ \ 729\\\underline{\ \ \ \ \ |\ 729}\\.\qquad\ \ \ 0$

$\sqrt{15129}=123$

$\begin{array}{c|c}15129&3\\5043&3\\1681&41\\41&41\\1\end{array}15129=3\cdot3\cdot41\cdot41=3^2\cdot41^2\\\\\sqrt{15129}=\sqrt{3^2\cdot41^2}=\sqrt3^2}\cdot\sqrt{41^2}=3\cdot41=123$

$Used:\\\\\sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b}\\\\\sqrt{a^2}=a\ for\ a\geq0$

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2 years ago

Ivan used coordinate geometry to prove that quadrilateral EFGH is a square.

Thepotemich [5.8K]

Answer:

this is geometry?

Step-by-step explanation:

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2 years ago

What is the following product? RootIndex 3 StartRoot 16 x Superscript 7 Baseline EndRoot times RootIndex 3 StartRoot 12 x Supers

KengaRu [80]

Answer:

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }$

Step-by-step explanation:

Given

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9}$

Required

Find the products

From laws of indices;

$\sqrt[m]{a} * \sqrt[m]{b} = \sqrt[m]{a*b}$

So;

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16x^7 * 12x^9}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16* x^7 * 12 * x^9}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^7 * x^9}$

From laws of indices

$a^m * a^n = a^(m+n); So,$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{7+9}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{16*12* x^{16}}$

Expand 16 * 12

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4*4*4*3* x^{16}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = \sqrt[3]{4^3 *3* x^{16}}$

From laws of imdices

$a^{\frac{1}{m}} = \sqrt[m]{a}$

So;

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^3 *3* x^{16}})^{\frac{1}{3}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4^{3*{\frac{1}{3}}} *3^{{\frac{1}{3}}}* x^{16*{\frac{1}{3}}}})$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{\frac{16}{3}}}$

Divide 16 by 3 (Write as ,mixed number)

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5\frac{1}{3}}}$

Split mixed numbers

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5+\frac{1}{3}}}$

Apply law of indices

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 *3^{{\frac{1}{3}}}* x^{5}*{x ^\frac{1}{3}}}$

Reorder

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 * x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*3^{{\frac{1}{3}}}*{x ^\frac{1}{3}}}$

Apply law of indices

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3} *\sqrt[3]{x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3*x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}*\sqrt[3]{3x} }$

$\sqrt[3]{16x^7} * \sqrt[3]{12x^9} = {4 x^{5}\sqrt[3]{3x} }$

6 0

2 years ago