Write an equation of the line containing the given point and perpendicular to the given line:
(7,- 4); 9x+7y=4

Step-by-step explanation:

The equation of a straight line can be represented in the slope intercept form as

y = mx + c

Where

m = slope = (change in the value of y in the y axis) / (change in the value of x in the x axis)

The equation of the given line is

9x+7y=4

7y = - 9x + 4

y = -9x/7 + 4/7

Comparing with the slope intercept form, slope = - 9/7

If the line passing through the given point is perpendicular to the given line, it means its slope is the negative reciprocal of the slope of the given line.

Therefore, the slope of the line passing through (7,-4) is 7/9

To determine the intercept, we would substitute m = 7/9, x = 7 and y = -4 into y = mx + c. It becomes

- 4 = 7/9×7 + c = 49/9 + c

c = - 4 - 49/9 = -85/9

The equation becomes

y = 7x/9 - 85/9

Step-by-step explanation:

any eq. of line perpendicular to 9x+7y=4 is

7x-9y=a

it passes  through (7,-4)

7(7)-9(-4)=a

49+36=a

a=85

reqd. eq. is 7x-9y=85

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

both graphs increase on their domain

neither graph has a y-intercept

step-by-step explanation:

the following points can be visualized easily from the graph