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3 years ago

Which expression represents the area of the rectangle (2x+7)(4x+5)?

Mathematics

1 answer:

zhenek [66]3 years ago

3 0

So we'll do FOIL (first, outer, inner, last)

The equation is (4x+5)(2x+7)

First, 4x·2x=8x²

Outer, 4x·7=28x

Inner, 5·2x=10x

Last, 5·7=35

Add like terms and you get:

8x²+38x+35

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ILL MARK BRAINIEST IF YOU DO THIS RIGHT!!!

Arlecino [84]

First let’s find the equation:
a = (42-21)/(26-13)
a= 21/13
y-y1 = a(x-x1)
y-21=(21/13)(x-13)
y-21=(21/13)x - 21
y=(21/13)x-21+21
y=(21/13)x

Now we will find the other point:
X = 2 >> y =(21/13)*2 = 3.23 (not)
X = 3 >> y =(21/13)*3 = 4.84 (not)
X = 4 >> y =(21/13)*4 = 6.46 (not)

X = 5 >> y =(21/13)*5 = 8.07 (yes) find the answer

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3 years ago

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4 years ago

Write an equation in point-slope form of the line that passes through (-1, – 4) with slope -2.

Rom4ik [11]

Y+4=-2(x+1)

This is the point slope form (y-y1) =m(x-x1)

I hope it will helped!

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3 years ago

For the following telescoping series, find a formula for the nth term of the sequence of partial sums {Sn}. Then evaluate limn→[

Ivenika [448]

Answer:

The following are the solution to the given points:

Step-by-step explanation:

Given value:

$1) \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\2) \sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

Solve point 1 that is $\sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2}\\\\$ :

when,

$k= 1 \to s_1 = \frac{1}{1+1} - \frac{1}{1+2}\\\\$

$= \frac{1}{2} - \frac{1}{3}\\\\$

$k= 2 \to s_2 = \frac{1}{2+1} - \frac{1}{2+2}\\\\$

$= \frac{1}{3} - \frac{1}{4}\\\\$

$k= 3 \to s_3 = \frac{1}{3+1} - \frac{1}{3+2}\\\\$

$= \frac{1}{4} - \frac{1}{5}\\\\$

$k= n^ \to s_n = \frac{1}{n+1} - \frac{1}{n+2}\\\\$

Calculate the sum $(S=s_1+s_2+s_3+......+s_n)$

$S=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.....\frac{1}{n+1}-\frac{1}{n+2}\\\\$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{n+1}-\frac{1}{n+2}\\\\$

When $s_n \ \ dt_{n \to 0}$

$=\frac{1}{2}-\frac{1}{5}+\frac{1}{0+1}-\frac{1}{0+2}\\\\=\frac{1}{2}-\frac{1}{5}+\frac{1}{1}-\frac{1}{2}\\\\= 1 -\frac{1}{5}\\\\= \frac{5-1}{5}\\\\= \frac{4}{5}\\\\$

$\boxed{\text{In point 1:} \sum ^{\infty}_{k = 1} \frac{1}{k+1} - \frac{1}{k+2} =\frac{4}{5}}$

In point 2: $\sum ^{\infty}_{k = 1} \frac{1}{(k+6)(k+7)}$

when,

$k= 1 \to s_1 = \frac{1}{(1+6)(1+7)}\\\\$

$= \frac{1}{7 \times 8}\\\\= \frac{1}{56}$

$k= 2 \to s_1 = \frac{1}{(2+6)(2+7)}\\\\$

$= \frac{1}{8 \times 9}\\\\= \frac{1}{72}$

$k= 3 \to s_1 = \frac{1}{(3+6)(3+7)}\\\\$

$= \frac{1}{9 \times 10} \\\\ = \frac{1}{90}\\\\$

$k= n^ \to s_n = \frac{1}{(n+6)(n+7)}\\\\$

calculate the sum: $S= s_1+s_2+s_3+s_n\\$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(n+6)(n+7)}\\\\$

when $s_n \ \ dt_{n \to 0}$

$S= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{(0+6)(0+7)}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}....+\frac{1}{6 \times 7}\\\\= \frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{42}\\\\=\frac{45+35+28+60}{2520}\\\\=\frac{168}{2520}\\\\=0.066$

$\boxed{\text{In point 2:} \sum ^{\infty}_{k = 1} \frac{1}{(n+6)(n+7)} = 0.066}$

8 0

3 years ago

Students make 71.5 ounces of liquid soap for a craft fair. They put the soap in 5.5 ounce bottles

Oxana [17]

Answer:

13 bottles of soap

Step-by-step explanation:

71.5/5.5=13

5 0

3 years ago