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umka21 [38]
3 years ago
5

Which value is an output of the function? O-6 O-2 O 4 O7

Mathematics
1 answer:
Black_prince [1.1K]3 years ago
3 0
The answer is the second one
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Solve the system of equations:<br> y= 2x - 5<br> y=x^2-5<br><br> Apex
arlik [135]

Answer:

A is the answer

Step-by-step explanation:

(0, -5):

-5 = 2(0) - 5 >>>   -5 = -5

-5 = (0)^2 - 5 >>>   -5 = -5

(2, -1):

-1 = 2(2) - 5 >>>   -1 = 4 - 5 >>>   -1 = -1

-1 = (2)^2 - 5 >>>   -1 = 4 - 5 >>>   -1 = -1

4 0
3 years ago
Evaluate 156 divided by(8x+14)when x=8 i am only in 6th grade be nice
VikaD [51]
So first we have to find out what the bottom number is so

8x+14
subsitute 8 for x
8(8)+14
64+14
78

so 156/78
an easy way to do this is to factor it out and find the 'ones' like 4/8=1/2 times 4/4 (4/4=1 and can be canceled out)
*=times
156=2*2*3*13
78=2*3*13

so (2*2*3*13)/(2*3*13)
2/1 times (2*3*13)/(2*3*13)
we can cross out the (2*3*13) and get
2 as the answer
6 0
3 years ago
Read 2 more answers
The derivative of the function f is given by f′(x)=−3x+4 for all x, and f(−1)=6. Which of the following is an equation of the li
Viefleur [7K]

Answer:

The equation of the line tangent to the graph of f at x = -1 is y = 7\cdot x +13.

Step-by-step explanation:

From Analytical Geometry we know that the tangent line is a first order polynomial, whose form is defined by:

y = m\cdot x + b (1)

Where:

x - Independent variable, dimensionless.

y - Dependent variable, dimensionless.

m - Slope, dimensionless.

b - Intercept, dimensionless.

The slope of the tangent line at x = -1 is:

f'(x) = -3\cdot x +4 (2)

f'(-1) = -3\cdot (-1) +4

f'(-1) = 7

If we know that m = 7, x = -1 and y = 6, then the intercept of the equation of the line is:

b = y-m\cdot x

b = 6-(7)\cdot (-1)

b = 13

The equation of the line tangent to the graph of f at x = -1 is y = 7\cdot x +13.

3 0
3 years ago
IF SOMEONE COULD HELP ME WITH THIS THAT WOULD BE GREAT 
evablogger [386]
Prove
 m∠5+m∠2+m∠6=180°

we have that
m∠1+m∠2+m∠3=180°   equation I

m∠1=m∠5  -------- > in<span>ternal alternate angles  equation 2
</span>
and

m∠6=m∠3-------- > internal alternate angles    equation 3

<span>I substitute  equation 2 and equation 3 in equation I</span>

m∠1+m∠2+m∠3=180°-------- > m∠5+m∠2+m∠6=180°------- > is ok
6 0
3 years ago
Enter the equation in slope-intercept form. Then graph the line described by the equation.
disa [49]

Answer:

2

4

6

8 10

ytyrfhkygzsdddddd

7 0
3 years ago
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