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VLD [36.1K]
3 years ago
7

13) (-2. – 5v).- (-4v – 2) Can you help a girl out

Mathematics
1 answer:
irga5000 [103]3 years ago
8 0

Answer:

If simplifed, it'll be -v

Step-by-step explanation:

You might be interested in
2.8 = (12*9.8 - 8/3 V²) ÷ 12 find V
Korvikt [17]

Answer:5.6

Step-by-step explanation:2.8×12=12×9.8-8/3v^2

33.6=117.6-8/3v^2

8/3v^2=117.6-33.6

8/3v^2=84

8v^2=84×3

8v^2=252

V^2=31.5

V=√31.5

V=5.6125

V~5.6

8 0
3 years ago
Rewrite the polynomial in the form ax^2 + bx + C and then identify the values of a, b, and c.
sergejj [24]

Answer:

a = 5\\\\b = -1\\\\c=1

Step-by-step explanation:

Given

-x+5x^2+1

Required

Rewrite and identify a, b and c

The required form is:

ax^2 + bx + c

Equate both expressions

ax^2 + bx + c = -x+5x^2+1

Rewrite as:

ax^2 + bx + c = 5x^2-x+1

Compare the expressions on either sides

a = 5\\\\b = -1\\\\c=1

8 0
3 years ago
Inequality question plz help
Serjik [45]

Answer:

C. x<18

Step-by-step explanation:

If you substituted the inequalities with 5, the logical one is x<18.

Especially since 5 is less than 18, it isn;t greater than or less than 2,9, or 42.

4 0
3 years ago
What is the equation of the quadratic graph with a focus of (1, 1) and a directrix of y = −1?
JulsSmile [24]
Alright,
directix is y=something so it opens down or up

we use
(x-h)²=4p(y-k)
the vertex is (h,k)
and p is distance from focus to vertex
if focus is above directix, p is positive
if focus is below directix, p is negative

so we gts

focus=(1,1)
directix is y=-1
1>-1
focus is above

oh, vertex is in middle of focus and directix
so
beteeen (1,1) and y=-1 is, hmm
that is a distance of 2 vertically
2/2=1
1 down from (1,1) is (1,0)
vertex is (1,0)
p=1

so
(x-1)²=4(1)(y-0)
solving for y to get into f(x)=something form
(x-1)²=4y
y=1/4(x-1)²
f(x)=1/4(x-1)²

4th option
6 0
3 years ago
Help needed....
LenKa [72]
The answer is C) y = 4x - 3
B and D can be eliminated by substituting 0 for x in each...the result is not -3
Now you have a 50/50 chance of getting the question correct
5 0
3 years ago
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