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Sav [38]
3 years ago
6

Which equation represents the line that is perpendicular to y = 1/4 and passes through (-6,-9)?

Mathematics
2 answers:
sergey [27]3 years ago
8 0

Answer:

I’m guessing D

Step-by-step explanation:

As the perpendicular would be -4 the negative reciprocal. I just worked out the whole equation including the point it passes.

AlekseyPX3 years ago
6 0

Answer:

-6

Step-by-step explanation:

You might be interested in
Two years ago, Rita was three times older than Cheryl. In 3 years, Rita will be twice older than Cheryl. How old are the girls n
Brrunno [24]

Answer:

Cheryl's age = x = 7 years

Rita's age = y = 17 years

Step-by-step explanation:

Let

Cheryl's age = x

Rita's age = y

Two years ago, Rita was three times older than Cheryl

(y - 2) = 3(x - 2)

y - 2 = 3x - 6

y = 3x - 6 + 2

= 3x - 4

y = 3x - 4

In 3 years, Rita will be twice older than Cheryl

(y + 3) = 2(x + 3)

y + 3 = 2x + 6

y = 2x + 6 - 3

= 2x + 3

y = 2x + 3

Equate both equations

3x - 4 = 2x + 3

Collect like terms

3x - 2x = 3 + 4

x = 7 years

Substitute x = 7 into

y = 2x + 3

= 2(7) + 3

= 14 + 3

= 17

y = 17 years

Cheryl's age = x = 7 years

Rita's age = y = 17 years

5 0
3 years ago
Make a tree diagram to show sample space. Then select the total number of outcomes.
Margaret [11]

Answer:

B. 10

Step-by-step explanation:

The first choice has 2 options: square or circle

The second choice has 5 options: orange, blue, yellow, red, or green.

2 x 5 = 10 total outcomes

7 0
3 years ago
Please help
Vlada [557]

We know that

When we have

f(x)=ax^2 +bx+c

Here ,

a is the leading coefficient

c is the constant term

so, we can compare it with

f(x)=5x^2-7x+6

so, we get

a=5,c=6

so,

leading coefficient is 5

constant term is 6......................Answer

4 0
4 years ago
Write the number in standard notation.
viva [34]
<span>1.14×10^3 = 1.14 x 1,000 = 1, 140

answer
</span><span>1,140</span>
7 0
3 years ago
Can someone please help me please!!!!!!!!!
Alla [95]
What do they mean by describe the diagram
7 0
3 years ago
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