All categories

3 years ago

What is the x-intercept of y=40x+70

Mathematics

1 answer:

AysviL [449]3 years ago

8 0

At th x-intercept y=o

∴40x + 70 = 0
40x = -70
x = -70/40
x = -1.75

x intercepr = (-1.75,0)

You might be interested in

What is the sum of 0.09 and 0.22

iren [92.7K]

I think the answer is 0.31!

7 0

3 years ago

Read 2 more answers

Put the equation in standard form with a positive x coefficient. 2y+5=-4(x+2)

umka21 [38]

The standard form of equation is: 4x+2y = -13

Step-by-step explanation:

the standard form of a linear equation in two variables is given by:

$ax+by = c$

Given equation is:

$2y+5=-4(x+2)$

Distributing -4 into the round bracket

$2y+5 = -4x-8$

Adding 4x on both sides

$4x+2y+5 = -4x-8+4x\\4x+2y+5 = -8$

subtracting 5 from both sides

$4x+2y+5-5 = -8-5\\4x+2y = -13$

The standard form of equation is: 4x+2y = -13

Keywords: Linear equation, variables

Learn more about linear equations at:

#LearnwithBrainly

6 0

3 years ago

doctors use gallium in scanning some patients for diseases the table shows the amount of gallium in mg remaining in a patients s

Blababa [14]

Step-by-step explanation:

no sn as. kojsshsusishheidhsusjjsbsbshsbsjs7ehsjsnkosvsjsishshshsjdjjddyfygyvrcudjdhdudhsjsudhdjs

7 0

3 years ago

Read the photo and it will tell the question

likoan [24]

Answer:

Step-by-step explanation:

u must add up the coordinates

6 0

4 years ago

Find the sum of the geometric series 512+256+ . . .+4

mario62 [17]

$\bf 512~~,~~\stackrel{512\cdot \frac{1}{2}}{256}~~,~~...4$

so, as you can see above, the common ratio r = 1/2, now, what term is +4 anyway?

$\bf n^{th}\textit{ term of a geometric sequence}\\\\a_n=a_1\cdot r^{n-1}\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\a_n=+4\end{cases}$

$\bf 4=512\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{4}{512}=\left( \cfrac{1}{2} \right)^{n-1}\\\\\\\cfrac{1}{128}=\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{1}{2^7}=\left( \cfrac{1}{2} \right)^{n-1}\implies 2^{-7}=\left( 2^{-1}\right)^{n-1}\\\\\\(2^{-1})^7=(2^{-1})^{n-1}\implies 7=n-1\implies \boxed{8=n}$

so is the 8th term, then, let's find the Sum of the first 8 terms.

$\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\n=8\end{cases}$

$\bf S_8=512\left[ \cfrac{1-\left( \frac{1}{2} \right)^8}{1-\frac{1}{2}} \right]\implies S_8=512\left(\cfrac{1-\frac{1}{256}}{\frac{1}{2}} \right)\implies S_8=512\left(\cfrac{\frac{255}{256}}{\frac{1}{2}} \right)\\\\\\S_8=512\cdot \cfrac{255}{128}\implies S_8=1020$

7 0

4 years ago