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belka [17]
3 years ago
9

Which expression of the gcf of the terms of the polynomial 16x^3+28x^5y

Mathematics
1 answer:
strojnjashka [21]3 years ago
4 0
1. Determine the prime factorization of 60. 
<span>a.2 • 3^2 • 5 </span>
<span>b.2^2 • 3 • 5 </span>
<span>c.3^2 • 7 </span>
<span>d.2^3 • 7 </span>

<span>2. Find the GCF of the terms of the polynomial. 16x^3 + 28x^5y </span>
<span>a.112x^5y </span>
<span>b.4x^5y </span>
<span>c.4x^3 </span>
<span>d.8x^5 </span>

<span>3. Simplify. 12n^2m^4/24^4m^3 </span>
<span>a.3n^2/7m </span>
<span>b.3n^2/m </span>
<span>c.3m/7n^2 </span>
<span>d.3n^6m^7/7 </span>

<span>4. Divide. (3a^3 + 6a^2) ÷ 3a </span>
<span>3a^3 + 6a^2 − 3a </span>
<span>3a^3 + 6a^2 + 3a </span>
<span>a^2 + 2a </span>
<span>a^2 + 6a^2 </span>
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Vsevolod [243]

Answer:

Part 1) x=14\sqrt{2}\ units

Part 2) y=14\ units

Part 3) z=14\sqrt{3}\ units

Step-by-step explanation:

see the attached figure with letters to better understand the problem

step 1

In the right triangle ABC

we know that

The triangle ABC is a 45^o-90^o-45^o

so

Is an isosceles right triangle

The legs are equal

therefore

AC=AB

x=14\sqrt{2}\ units

step 2

Find the length side BC

Applying the Pythagorean Theorem]

BC^2=AB^2+AC^2

BC^2=(14\sqrt{2})^2+(14\sqrt{2})^2

BC^2=784\\BC=28\ units

step 3

Find the value of y

In the right triangle BCD

sin(30^o)=\frac{BD}{BC} ----> by SOH (opposite side divided by the hypotenuse)

sin(30^o)=\frac{y}{28}

Remember that

sin(30^o)=\frac{1}{2}

so

\frac{y}{28}=\frac{1}{2}\\\\y=14\ units

step 4

Find the value of z

In the right triangle BCD

cos(30^o)=\frac{DC}{BC} ----> by CAH (adjacent side divided by the hypotenuse)

cos(30^o)=\frac{z}{28}

Remember that

cos(30^o)=\frac{\sqrt{3}}{2}

so

\frac{z}{28}=\frac{\sqrt{3}}{2}\\z=14\sqrt{3}\ units

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