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Rainbow [258]
3 years ago
5

Factor 20x2 + 25x – 12x – 15 by grouping. 1. Group terms with common factors. 2. Factor the GCF from each group. 3. Write the po

lynomial as a product of binomials.
Mathematics
2 answers:
a_sh-v [17]3 years ago
8 0

Answer:

(5x - 3) • (4x + 5)

Step-by-step explanation:

RUDIKE [14]3 years ago
5 0
<span>Step 1:
</span>
<span> (((22•5x2) + 25x) - 12x) - 15)))

</span>
<span>The first term is, <span> <span>20x2</span> </span> its coefficient is <span> 20 </span>.
The middle term is, <span> +13x </span> its coefficient is <span> 13 </span>.
The last term, "the constant", is <span> -15



</span></span>
<span>step 2 above,  -12  and  25 
                     <span>20x2 - 12x</span> + 25x - 15

Step-4 : Add up the first 2 terms, pulling out like factors :
                    4x • (5x-3)
              Add up the last 2 terms, pulling out common factors :
                    5 • (5x-3)
Step-5 : Add up the four terms of step 4 :
                    (4x+5)  •  (5x-3)
             Which is the desired factorization</span>Final result :<span> (5x - 3) • (4x + 5) </span>
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3(x-4)

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What is the following product? Assume x&gt;/=0 ^3sqrt(x^2)*^sqrtx^3
RoseWind [281]

Answer:

x(\sqrt[12]{x^5} )

Step-by-step explanation:

We need to remember 2 rules when doing these:

1. \sqrt[n]{x^a} =x^{\frac{a}{n}}

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Using these 2 rules, we can simplify the product (steps shown below):

\sqrt[3]{x^2} *\sqrt[4]{x^3} \\=x^{\frac{2}{3}}*x^{\frac{3}{4}}\\=x^{\frac{2}{3}+\frac{3}{4}}\\=x^{\frac{17}{12}}\\=x^{\frac{12}{12}+\frac{5}{12}}\\=x(x^{\frac{5}{12}})\\=x(\sqrt[12]{x^5} )

Rearranging, we see that it is the third choice.

7 0
3 years ago
What is AE and the perimeter of BCDE? Round your answer to the nearest tenth.
Oliga [24]

Answer:

AE = 43.2 units

Perimeter = 229.2 units

Step-by-step explanation:

Let the side AE be 'x'.

Consider triangles AEB and ADC

Statements                                                     Reasons

1. ∠ ABE ≅ ∠ ACD                                        Right angles are congruent.

2. ∠A ≅ ∠A                                                   Common angle

Therefore, the two triangles are similar by AA postulate.

Now, for similar triangles, the ratio of their corresponding sides are also proportional to each other. Therefore,

\frac{AE}{AD}=\frac{AB}{AC}=\frac{BE}{DC}\\\\\frac{AE}{AE+ED}=\frac{AC-BC}{AC}

Now, plug in the given values and solve for 'x'. This gives,

\frac{x}{x+72}=\frac{88-55}{88}\\\\88x=33(x+72)\\\\88x=33x+2376\\\\88x-33x=2376\\\\55x=2376\\\\x=\frac{2376}{55}=43.2

Therefore, AE = 43.2 units

Now, from right angled triangle ABE,

(AB)^2+(BE)^2=(AE)^2...(Pythagoras\ Theorem)\\\\(33)^2+(BE)^2=(43.2)^2\\\\BE=\sqrt{(43.2)^2-(33)^2}=27.879

Similarly from right angled triangle ACD,

CD=\sqrt{AD^2-AC^2}\\\\CD=\sqrt{(72+43.2)^2-88^2}=74.344

Now, perimeter is the sum of all the sides of a figure. Therefore, the perimeter of BCDE is given as:

Perimeter = BE + ED + CD + BC

Perimeter = 27.879 + 72 + 74.344 + 55 = 229.223 ≈ 229.2 (Nearest tenth)

Therefore, the perimeter = 229.2 units

6 0
3 years ago
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Answer:

About $0.2708 per unit ' each can '.

5 0
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Read 2 more answers
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