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3 years ago

PLZ HELP ASAP RIGHT TRIANGLES

Mathematics

2 answers:

Galina-37 [17]3 years ago

4 0

6.02 because you divide
4.3 ÷ 4= 1.075
Now multiply 1.075 × 5.6 = 6.02

Phantasy [73]3 years ago

4 0

By similarity of triangles we have the following expression:
x / 4 = 5.6 / 4.3
Thus, by clearing the value of x from the expression dad, we have:
x = (5.6 / 4.3) * 4
Calculating x we have:
x = 5.209302326
Rounding off we have:
x = 5.21
Answer:
The value of x is:
x = 5.21

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a globe has a diameter of 24 inches. a smaller globe has a diameter of 18 inches. What is the surface are to volume ratio of the

faust18 [17]

Answer:

1 pi: 3pi

Step-by-step explanation:

Step 1: Formula of surface area and volume of sphere

Surface area of sphere = 4 x pi x r^2

Volume of sphere = <u>4</u> x pi x r^3

Step 2: Apply values in the formula

r = radius

radius = diameter/2

r=18/2 = 9

S.A = 4 x pi x 9^2

S.A = 324pi

Volume = <u>4</u> x pi x 9^3

Volume = 972pi

Step 3 : Show in ratio

Surface area : Volume

324pi : 972pi

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Sergeu [11.5K]

Answer:

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Step-by-step explanation:

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3 years ago

1) Given the function f(x) = 2x -3, evaluate f(3).

igomit [66]

Answer:

1) $f(3) = 3$

2) $x = 2$

Step-by-step explanation:

Given:

$f(x) = 2x -3$

Evaluating $f(3)$ :

$f(3) = 2(3) -3 \\ f(3) = 6 -3 \\ f(3) = 3$

Solving for $f(x) = -1$ :

$2x -3 = 1 \\ 2x -3 +3 = 1 +3 \\ 2x = 4 \\ \frac{2x}{2} = \frac{4}{2} \\ x = 2$

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3 years ago

Read 2 more answers

SOMEBODY HELP ME PLS!

babunello [35]

Answer:

To find the value of y from a given value of x, find the position of x on the x-axis, then trace vertically until you meet the line. Once you meet the line, trace horizontally to the y-axis to find the corresponding value of y.

To find the value of x from a given value of y, find the position of y on the y-axis, then trace horizontally until you meet the line. Once you meet the line, trace vertically to the x-axis to find the corresponding value of x.

From inspection of the graph,

when x = 1, y = -1

when x = 0, y = 2

when x = 2, y = -4

Therefore,

$\begin{array}{| c | c |}\cline{1-2} x & y\\\cline{1-2} 1 & -1 \\\cline{1-2} 0 & 2 \\\cline{1-2} 2 & -4 \\\cline{1-2}\end{array}$