2 years ago

How many different triangles can you make if you are given these two lengths for sides?

Mathematics

1 answer:

balandron [24]2 years ago

8 0

Answer:

Step-by-step explanation:

The sides are equivalent enough to be a triangle, but only 1.

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X - 7y - 10 - x - 7y + 10 x - 7y + 10 Find the quotient. (5x - 35y + 50) ÷ 5

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Divide each term by 5, looking at the coefficients. The answer is x - 7y + 10

Step-by-step explanation:

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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

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

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

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

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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