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2 years ago

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Which number can each term of the equation be multiplied by to eliminate the fractions before solving? 6 – x + = 6 minus StartFr

action 3 Over 4 EndFraction x plus StartFraction 1 Over 3 EndFraction equals StartFraction one-half EndFraction x plus 5.x + 5

1 answer:

zheka24 [161]2 years ago

7 0

Answer:

We need to multiply 12 to each term to eliminate fractions.

Step-by-step explanation:

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Factor p^2−6−q·(p^2−6)^2

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Answer:

(p^2−6) (1-q(p^2-6))

Step-by-step explanation:

p^2−6−q·(p^2−6)^2

Put parentheses around the P^2-6 at the beginning of the expression

(p^2-6) -q (p^2−6)^2

Factor out (p^2−6)

(p^2−6) (1-q(p^2-6))

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Evaluate S5 for 300 + 150 + 75 + … and select the correct answer below. 18.75 93.75 581.25 145.3125

Savatey [412]

we have that

$300 + 150 + 75 +...$

Let

$a1=300\\ a2=150\\ a3=75$

we know that

$\frac{a2}{a1} =\frac{150}{300} \\\\ \frac{a2}{a1}=0.5 \\ \\ a2=a1*0.50$

$\frac{a3}{a2} =\frac{75}{150} \\\\ \frac{a3}{a2}=0.5 \\ \\ a3=a2*0.50$

so

$a(n+1)=an*0.50$

Is a geometric sequence

Find the value of $a4$

$a(4)=a3*0.50$

$a(4)=75*0.50$

$a(4)=37.5$

Find the value of $a5$

$a(5)=a4*0.50$

$a(5)=37.5*0.50$

$a(5)=18.75$

Find $S5$

$S5=a1+a2+a3+a4+a5\\ S5=300+150+75+37.5+18.75\\ S5=581.25$

therefore

the answer is

$581.25$

Alternative Method

Applying the formula

$S_n=\frac{a_1 (1-r^n)}{1-r} \\\\a_1=300 \\ r=\frac{1}{2}\\\\ S_5=\frac{300(1-(\frac{1}{2})^5)}{1-\frac{1}{2}}\\\\=\frac{300(1-\frac{1}{32})}{\frac{1}{2}}\\\\=\frac{300 \times \frac{31}{32}}{\frac{1}{2}}\\\\=\frac{75 \times \frac{31}{8}}{\frac{1}{2}}\\\\=\frac{\frac{2325}{8}}{\frac{1}{2}}\\\\=\frac{2325}{8} \times 2\\\\=\frac{2325}{4}\\\\=581 \frac{1}{4}\\\\=581.25$

therefore

the answer is

$581.25$

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