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3 years ago

Can someone help me with this last one

Mathematics

1 answer:

Lena [83]3 years ago

8 0

The answer is C because the triangles are congruent by angle side angle theorem. Two sets of matching angles with a pair of matching sides in between.
when solving questions like these always skip the whole flip it over part. look at the measurements and the order they are in across the shape and find the pattern in the answer.

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A two-word phrase used to show division in a word problem.

fomenos

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3 years ago

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Jo is looking for a new part-time job as a plumber. She responds to a classified ad for a position that pays 44.9K. What would h

hoa [83]

Answer:

855.77

Step-by-step explanation:

44500/52 equals 855.77

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3 years ago

Help with inequalities

max2010maxim [7]

Answer:

$2(b - 9) \geqslant 26$

Step-by-step explanation:

difference of a number and 9 means to subtract

$b - 9$

twice that means multiply by 2

$2(b - 9)$

at least is greater than or equal to

$2(b - 9) \geqslant 26$

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3 years ago

Find the equation of the line with Slope = 6 and passing through (5,38). Write your equation in the form y =mx+b

laiz [17]

Hello.

We have:

A point that the line passes through
The slope of the line

Plug the values into the Point-Slope Formula:

$\mathrm{y-y_1=m(x-x_1)}$

$\mathrm{y-38=6(x-5)}$

$\mathrm{y-38=6x-30}$

$\mathrm{y=6x-30+38}$

$\Large\boxed{\sf{ \ y=6x+8}}$

I hope it helps.

Have a nice day.

$\boxed{imperturbability}$

5 0

2 years ago

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If g (x) = 1/x then [g (x+h) - g (x)] /h

lys-0071 [83]

Answer:

$\dfrac{-1}{x(x+h)}, h\ne 0$

Step-by-step explanation:

If $g(x) = \dfrac{1}{x}$ , then $g(x+h) = \dfrac{1}{x+h}$ . It follows that

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \cdot [g(x+h) - g(x)] \\&= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\end{aligned}$

Technically we are done, but some more simplification can be made. We can get a common denominator between 1/(x+h) and 1/x.

$\begin{aligned} \\\frac{g(x+h)-g(x)}{h} &= \frac{1}{h} \left( \frac{1}{x+h} - \frac{1}{x} \right)\\&=\frac{1}{h} \left(\frac{x}{x(x+h)} - \frac{x+h}{x(x+h)} \right) \\ &=\frac{1}{h} \left(\frac{x-(x+h)}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{x-x-h}{x(x+h)}\right) \\ &=\frac{1}{h} \left(\frac{-h}{x(x+h)}\right) \end{aligned}$

Now we can cancel the h in the numerator and denominator under the assumption that h is not 0.

$= \dfrac{-1}{x(x+h)}, h\ne 0$

5 0

3 years ago