What are the angle names given in the picture?

Mathematics

1 answer:

alisha [4.7K]3 years ago

4 0

Answer:

tbh there are quite a lot of possibilities

I'll just name the properties but not show which angle applies to it cos that's too time consuming

Step-by-step explanation:

vertically opposite angles

interior / exterior alternate angles

corresponding angles

angles on a straight line are supplementary

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In the function y=5x^2-2, what effect does the number 5 have on the graph, as compared to the graph of y=x^2?

allochka39001 [22]

General Idea:

The Rules for Transformations of Functions are given below:

If f(x) is the original function, a > 0 and c > 0; Then

$f(x)+k \; \; shift \; f(x) \; UPWARD \; k \; units\\\\f(x)-k \; \; shift \; f(x) \; DOWNWARD \; k \; units\\\\f(x+h) \; \; shift \; f(x) \; LEFT \; h \; units\\\\f(x-h) \; \; shift \; f(x) \; RIGHT \; h \; units\\\\-f(x) \; \; REFLECT \; f(x) \; in \; the \; x-axis\\\\f(-x) \; REFLECT \; f(x) \; in \; the \; y-axis\\\\a \cdot f(x), \; a>1 \; STRETCH \; f(x) \; vertically \; by \; a \; factor \; of \; a\\\\a \cdot f(x), \; 0$

$f(ax), \; a>1 \; SHRINK \; f(x) \; horizontally \; by \; a \; factor \; of \;\frac{1}{a} .\\\\f(ax), \; 0$

Applying the concept:

In the function y=5x^2-2, the effect that the number 5 have on the graph, as compared to the graph of y=x^2 is given below:

C.it stretches the graph vertically by a factor of 5

4 0

3 years ago

Read 2 more answers

1 Point

I am Lyosha [343]

Option C

The ratio for the volumes of two similar cylinders is 8 : 27

<h3><u>Solution:</u></h3>

Let there are two cylinder of heights "h" and "H"

Also radius to be "r" and "R"

$\text { Volume of a cylinder }=\pi r^{2} h$

Where π = 3.14 , r is the radius and h is the height

Now the ratio of their heights and radii is 2:3 .i.e

$\frac{\mathrm{r}}{R}=\frac{\mathrm{h}}{H}=\frac{2}{3}$

<em><u>Ratio for the volumes of two cylinders</u></em>

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\frac{\pi r^{2} h}{\pi R^{2} H}$

Cancelling the common terms, we get

$\frac{\text {Volume of cylinder } 1}{\text {Volume of cylinder } 2}=\left(\frac{\mathrm{r}}{R}\right)^{2} \times\left(\frac{\mathrm{h}}{\mathrm{H}}\right)$