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3 years ago

Find the greatest common factor

Mathematics

1 answer:

timama [110]3 years ago

4 0

answers to 1 - 4:

1. GCF: 2

2. GCF: 3

3. GCF : 7

4. GCF: 3

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Caroline puts $400 into a savings account that earns 6% annually. Write a function that represents this situation, where t is th

DaniilM [7]

Answer: 19 years

Step-by-step explanation:

Given

Amount invested $\$400$

Rate of interest $R=6\%$

after 1-year amount becomes

$A=400\left (1+6\%\right)\\A=400\left (1+0.06\right)\\A=400\times 1.06\\A=\$424$

After n years it is

$A=400\left (1.06\right)^n$

Sum accumulated is $\$1200$

$\Rightarrow 1200=400\left(1.06\right)^n\\\Rightarrow 3=\left(1.06\right)^n\\\text{Taking log both sides}\\\Rightarrow n=\dfrac{\ln 3}{\ln 1.06}\\\\\Rightarrow n=18.85\approx 19\ \text{years}$

5 0

3 years ago

Please help me ASAP!!! I think it's B but I don't know for sure

Katarina [22]

Answer: it's B

Step-by-step explanation:

you just have to use the AG and work backwards to find AD

7 0

3 years ago

Find the sales tax. Sales Tax nbsp Selling Price nbsp nbsp Rate of Sales Tax nbsp nbsp Sales Tax nbsp $70.00 6% ? The sales t

Lena [83]

Answer:

Sales tax is $4.20

Step-by-step explanation:

70*6/100=4.2

5 0

3 years ago

Factor the expression 5x^2 -22x -15

Murljashka [212]

$5x^2-22x-15=5x^2\underbrace{-25x+3x}_{-22x}-15\\\\=(5x^2-25x)+(3x-15)=5x(x-5)+3(x-5)\\\\=\boxed{(x-5)(5x+3)}$

3 0

3 years ago

1) find the equation of the line parallel to

In-s [12.5K]

Answer:

1) The equation of the line parallel to x-5y=6 and through (4,-2) is 5y = x -14

2) The equation of the line perpendicular to y= -2/5x + 3 and through (2,-1) is 2y = 5x -12

Solution:

1) find the equation of the line parallel to x-5y=6 and through (4,-2).

Given, line equation is x – 5y = 6

We have to find the line equation that is parallel to given line and passing through the point (4, -2)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-1}{-5}=\frac{1}{5}$

Now, we know that, slope of parallel lines are equal.

So, slope of required line is 1/5 and it passes through (4, -2)

Now, using point slope form

$y-y_{1}=m\left(x-x_{1}\right) \text { where } m \text { is slope and }\left(x_{1}, y_{1}\right) \text { is point on line. }$

$y-(-2)=\frac{1}{5}(x-4) \rightarrow 5(y+2)=x-4 \rightarrow 5 y+10=x-4 \rightarrow x-5 y=14$

Hence, the line equation is 5y = x -14

2) find the equation of the line perpendicular to y= -2/5x + 3 and through (2,-1)

$\text { Given, line equation is } y=-\frac{2}{5} x+3 \rightarrow 5 y=-2 x+15 \rightarrow 2 x+5 y=15$

We have to find the line equation that is perpendicular to given line and passing through the point (2, -1)

Now, let us find slope of the given line.

$\text { Slope }=\frac{-x \text { coefficient }}{y \text { coefficient }}=\frac{-2}{5}=-\frac{2}{5}$

Now, we know that, product of slopes of perpendicular lines equals to -1.

So, slope of required line $\times$ slope of given line = -1

slope of required line = $-1 \times \frac{5}{-2}=\frac{5}{2}$

And it passes through (2, -1)

Now, using point slope form

$\begin{array}{l}{\text { Line equation is } y-(-1)=\frac{5}{2}(x-2) \rightarrow 2(y+1)=5(x-2)} \\\\ {\rightarrow 2 y+2=5 x-10 \rightarrow 5 x-2 y=12}\end{array}$

Hence, the line equation is 2y = 5x -12

4 0

3 years ago