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3 years ago

The larger of two consecutive odd integers is three times the smaller. What is their sum?

Mathematics

1 answer:

Darya [45]3 years ago

4 0

X=y+2
x=3y
2x=4y+2
4y+2=2(y+2)
4y+2=2y+4
2y+2=4
2y=2
y=1
x=3
1+3=4

Their sum is 4

Hope this helps :)

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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

$581.25$

6 0

3 years ago

Read 2 more answers

A 90% confidence interval for the average weight (in pounds) of Americans was found to be (185, 210). The point estimate for thi

Lynna [10]

Answer:

B. 12.5

Step-by-step explanation:

We have the lowe confidence interval = 185

The upper confidence interval = 210

Mean of X = (lower confidence + upper confidence interval)/2

Mean of X = 185 + 210/2

= 197.5

The margin of error = the upper confidence interval - mean of X

= 210-197.5

= 12.5

3 0

3 years ago

20 PTSS !!! PLEASE HELPP !!

butalik [34]

Because it is extremely hard to find the area of this figure all together, it would be in our best interest to split this figure up into three different pieces: the two horizontal rectangles, and the verticle rectangle. We can find the area of all three and add them up. Be aware that there are two different ways that you can break this figure up, As shown in the attachments. I will be using the first image (the one with the tall horizontal rectangles, NOT the almost-squares).

So, we see that we have enough information to solve for the area of the left-most rectangle. Area = lw. 10 x 4 = 40, so the area is 40. Next, we have to notice, that the horizontal rectangles are also the same, so both of the areas of the two horizontal rectangles are 40.

Now, we can find the middle rectangle. We know that the length of the entire thing is 18, but it is taken up by 8 (4+4) of the horizontal triangles, so 18-8=10, so the length Is 10. We also know that the height of the horizontal rectangles is 10, so 10-3=7. Our dimensions for the rectangle are 10x7 or 70 square units. If we add them all together, 40+40+70=150.

The area is 150 square units