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slega [8]
3 years ago
8

Consider the line 6x-7y = 7.

Mathematics
1 answer:
Maurinko [17]3 years ago
8 0

Answer:

Consider the line 6x-7y = 7.

What is the slope of a line parallel to this line?

What is the slope of a line perpendicular to this line?

Step-by-step explanation:

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kate has the following scores on her math test this quarter 89,79,83, and 94.if she want an average of 85 ,what musther score be
Georgia [21]
She must score an 80.

85*5= 425
425-89-79-83-94= 80
8 0
1 year ago
Find the equation of the line that contains the given point and the given slope. Write the equation in slope-intercept form.
stealth61 [152]

The slope-point form of a line:

y-y_0=m(x-x_0)

The slope-intercept form of a line:

y=mx+b

1.

m=6,\ (4,\ 1)\to x_0=4,\ y_0=1

Substitute

y-1=6(x-4)\qquad|\text{use distributive property}\\\\y-1=6x-24\qquad|\text{add 1 to both sides}\\\\\boxed{y=6x-23}

2.

m=-5,\ (6,\ -3)

Substitute

y-(-3)=-5(x-6)\qquad|\text{use distributive property}\\\\y+3=-5x+30\qquad|\text{subtract 5 from both sides}\\\\\boxed{y=-5x+24}

3.

m=-\dfrac{1}{2},\ (-8,\ 2)\\\\y-2=-\dfrac{1}{2}(x-(-8))\\\\y-2=-\dfrac{1}{2}(x+8)\\\\y-2=-\dfrac{1}{2}x-4\qquad|\text{add 2 to both sides}\\\\\boxed{y=-\dfrac{1}{2}x-2}

4.

m=0,\ (-7,\ -1)\\\\y-(-1)=0(x-(-7))\\\\y+1=0\qquad|\text{subtract 1 from both sides}\\\\\boxed{y=-1}

3 0
3 years ago
Is x + y = 18 linear or non linear?
r-ruslan [8.4K]
A linear equation has 1 for all exponents on the placeholders
aka the equation is first degree (aka the hightest exponent on placeholders=1) so
it is linear
7 0
2 years ago
Read 2 more answers
Can someone pls help me?
nignag [31]

Step-by-step explanation:

ofc , so that is that and the other is that

4 0
3 years ago
Determine whether the improper integral converges or diverges, and find the value of each that converges.
shusha [124]

Answer:convergent

Step-by-step explanation:

Given

Improper Integral I is given as

I=\int_{-\infty}^{0}1000e^xdx

I=1000=\int_{-\infty}^{0}e^xdx

integration of e^x is e^x

I=1000\times \left [ e^x\right ]^{0}_{-\infty}

I=1000\times I=\left [ e^0-e^{-\infty}\right ]

I=1000\times \left [ e^0-\frac{1}{e^{\infty}}\right ]

I=1000\times 1=1000

so the integration converges to 1000 units  

6 0
3 years ago
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