Use the information giving in the figure to find the length FH. If applicable, round your answer to the nearest whole number.

Mathematics

1 answer:

Dimas [21]3 years ago

5 0

9514 1404 393

Answer:

FH = 16

Step-by-step explanation:

ΔGHE is a 5-12-13 triangle.

ΔEHF is a 3-4-5 triangle with a scale factor of 4, so side FH is 4·4 = 16 units.

_____

(3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17) are all commonly used Pythagorean triples. The first two on the list are used in these triangles.

If you like, you can work out the numbers using the Pythagorean theorem, which tells you the square of the hypotenuse is equal to the sum of the squares of the other two sides.

For ΔEHG, that is ...

GE² = EH² + HG²

13² = EH² + 5²

EH² = 13² -5² = 169 -25 = 144

And for ΔEHF, ...

FE² = EH² +HF²

20² = 144 + HF²

400 -144 = HF² = 256

HF = √256 = 16

The length of side FH is 16 units.

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Multiply. Write your answer as a fraction in simplest form 9/10 x 2/3

Oduvanchick [21]

Answer:

3/5

Step-by-step explanation:

9/10 and 2/3 can cross cancel

3 goes into 9, 3 times

2 goes into 10, 5 times

they both go into themselves once

our new fractions are 3/5 and 1/1 which equals 3/5

8 0

3 years ago

Read 2 more answers

Greg is trying to solve a puzzle where he has to figure out two numbers, x and y. Three less than two-third of x is greater than

Fittoniya [83]

Inequation 1:

$\frac{2}{3}x-3 \geq y$

to plot the pairs (x, y) for which the inequation holds, draw the line $y=\frac{2}{3}x-3$

then pick a point in either side of the line. If that point is a solution of the inequation, than color that region of the line, if that point is not a solution, then color the other part of the line.

we do the same for the second inequation. Then the solution, is the region of the x-y axes colored in both cases.

inequation 2:

$y+ \frac{2}{3}x\ \textless \ 4$

$y\ \textless \ - \frac{2}{3} x+ 4
$

draw the lines

i) $y=\frac{2}{3}x-3$ use points (0, -3), (3, -1)

ii) $y=- \frac{2}{3} x+ 4$ use points ( 0, 4), (3, 2)

let's use the point P(3, 3) to see what region of the lines need to be coloured:

$\frac{2}{3}x-3 \geq y$ ;
$\frac{2}{3}(3)-3 \geq 3$
$2-3 \geq 3$ , not true so we color the region not containing this point

$y+ \frac{2}{3}x\ \textless \ 4$
$(3)+ \frac{2}{3}(3)\ \textless \ 4$
$3+ 5\ \textless \ 4$ not true, so we color the region not containing the point (3, 3)

The graph representing the system of inequalities is the region colored both red and blue, with the blue line not dashed, and the red line dashed.