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Maurinko [17]
3 years ago
8

I need the answer to this

Mathematics
1 answer:
aleksandr82 [10.1K]3 years ago
8 0

Answer:

50

Step-by-step explanation:

All three angles of a triangle will add up to 180. Since one angle is known, 90, the right angle, and another is known, 40, we can solve for the missing angle.

90 + 40 + x = 180

Add like terms.

130 + x = 180

Subtract 130 from both sides.

x = 50

The missing angle is 50.

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What common error(s) did this student make? What is the correct solution for x? Please help
Reika [66]

Answer:

10x - 5(x - 1) = 20

10x - 5x + 5 = 20

5x + 5 = 20

5x = 20 - 5

5x = 15

x = 15/5

x = 3

Step-by-step explanation:

7 0
3 years ago
Read 2 more answers
Consider the line y=5x -3 Find the equation of the line that is patellar to this line and passes through the point (-8, -3) Cons
Lynna [10]

Answer:

The equation of the line that is parallel to given line and passes through the point (-8, -3) is: y = 5x+37

Step-by-step explanation:

Given equation of line is:

y=5x-3

The general form of equation of line in slope-intercept form is written as:

y = mx+b

Here m(co-efficient of x) is the slope of the line and b is the y-intercept.

Comparing the given equation with the general form we get

m = 5

Two parallel lines have same slope so the slope of any line parallel to given line will also be 5.

Let m1 be the slope of required line parallel to y=5x-3

Then m1=5

Putting in general form

y = m_1x+b

y=5x+b

To find the value of b(y-intercept) the given point has to be put in the equation from which the line passes.

The point is (-8,-3)

-3 = 5(-8)+b\\-3 = -40+b\\b = -3+40\\b = 37

Putting the value of b and m1, we get

y = 5x+37

Hence,

The equation of the line that is parallel to given line and passes through the point (-8, -3) is: y = 5x+37

5 0
3 years ago
Can someone please help me:)
Blababa [14]
The correct answer is c
8 0
3 years ago
What is the slope of this graph? A. −14 B. 4 C.−4 D. 14
DiKsa [7]
What graph are you talking about:? can you give a pivture
6 0
4 years ago
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Last derivatives based problem. Help would be appreciated.
lapo4ka [179]

The derivative of f(x) is defined by the limit,

f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}h

In this case, f(x)=8x-x^2-4, so the derivative would be

f'(x)=\displaystyle\lim_{h\to0}\frac{(8(x+h)-(x+h)^2-4)-(8x-x^2-4)}h

Simplifying the numerator gives

\dfrac{(8x+8h-x^2-2xh-h^2-4)-(8x-x^2-4)}h=\dfrac{8h-2xh-h^2}h

The numerator and denominator share a common factor of h, which we can cancel because h\to0 means that we're considering h\neq0 so that \dfrac hh=1:

\dfrac{8h-2xh-h^2}h=8-2x-h

Then as h\to0, we're left with the derivative

f'(x)=8-2x

4 0
4 years ago
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