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Tresset [83]
3 years ago
10

Help! (pic above)

Mathematics
2 answers:
MissTica3 years ago
8 0

Step-by-step explanation:

(10x + 3) + 4x \\ (3 + 10x) + 4x \\ 3 + 10x + 4x \\ 3 + 14x \\  \\ option \: (a) \\ thank \: you

Ganezh [65]3 years ago
3 0

Answer:

line 1 to line 2 is c line 2 to line 3 is a

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Two numbers total 63 and have a difference of 31. Find the two numbers
Anastaziya [24]

Answer: 47 and 16

Step-by-step explanation:

- Make Two Equations

x + y = 63

x - y = 31

- Set one of the equations equal to one of the variables

x + y = 63

x = 31 + y

- Substitute the equation back into the other one

(31 + y) + y = 63

31 + 2y = 63

2y = 32

y = 16

- Substitute the answer back into the equation

x + y = 63

x + 16 = 63

x = 47

5 0
3 years ago
Read 2 more answers
Without using a calculator, match each expression to the correct point.<br><br> Point<br> Expression
lyudmila [28]

Answer:

Where is the photo?

Step-by-step explanation:

7 0
3 years ago
Bill takes a loan of $8,000.00 at a 14% simple interest rate for 7 years. How much interest will he have to pay after 6 years
mestny [16]
I believe this answer is 186.66
4 0
3 years ago
Want Brainliest? Get this correct , Which rule should be applied to reflect f(x) = x^3 over the y-axis?
Rashid [163]

Answer:

B

Step-by-step explanation:

When you flip over the y-axis you the x is multiplied by -1

8 0
3 years ago
P(x)=3x^2+12x+7 find coordinates of vertex and intercepts
kotegsom [21]
Te x intercepts are where the graph crosses the x axis or wher y=0
the y intercept is where the graph crosses the y axis or where x=0

to find vertex, here is a hack
the x coordinate of the vertex for an equation in form ax^b+bx+c=y is -b/2a

so
y=3x^2+12x+7
-b/2a=-12/2(3)=-12/6=-2
sub that back
y=3(-2)^2+12(-2)+7
y=3(3)-24+7
y=9-17
y=-8

vertex is (-2,-8)

intercepts
x intercept is where y=0

0=3x^2+12x+7
using quadratic formula
x=(-6-√15)/3 and (-6+√15)/3
xints at ((-6-√15)/3,0) and ((-6+√15)/3,0)

yint is where x=0
set x=0
y=7
yint is at y=7 or (0,7)


vertex at (-2,-8)
xints at x=\frac{-6+ \sqrt{15} }{3} and \frac{-6- \sqrt{15} }{3} or at the points ( \frac{-6+ \sqrt{15} }{3} , 0) and (\frac{-6- \sqrt{15} }{3},0)
yint at y=7 or at (7,0)
8 0
3 years ago
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