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2 years ago

Which set of line segments could create a right triangle?

Mathematics

2 answers:

Klio2033 [76]2 years ago

4 0

Answer:

the 3rd one got it right on mine

Step-by-step explanation:

AleksandrR [38]2 years ago

3 0

Answer:

A right triangle will satisfy Pythagoras Theorem:

c² = a² + b²; where c is the longest side.

The only segments satisfying this are

15, 36, 39

Step-by-step explanation:

You might be interested in

Can someone do this for me or help me

Elden [556K]

JM=12
JL= 14
MN=?
MK=?

VT= 11
UV= 9
RS=?
ST=?

GF=23
HF=20
GH=?
GE=?

M<1=?
M<2=?
M<3=?
M<4=?
M<5=?
M<6=?
M<7=?
M<8 = 90 degrees

WXZ = 34 degrees
WVZ=90 degrees
ZYW= 56 degrees

These are the only answers I knew, I’m sorry I couldn’t find the rest. If I do find more answers, I’ll comment them.

5 0

2 years ago

Emma had $2.40 to spend on seven pencils. After buying them she had $1.00. How much did each pencil cost?

OlgaM077 [116]

Answer:

Each pencil costs 20 cents

Step-by-step explanation:

2.40 - 1 = 1.40

1.40/7 = 0.2

8 0

3 years ago

Read 2 more answers

Please answer the picture i put! Thanks Sooo much.

kow [346]

The answer is A and B

5 0

2 years ago

Read 2 more answers

Drag the tiles to the boxes to form correct pairs. Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

Solving the expression

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

Solving the expression

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

Solving the expression

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

Solving the expression

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

Learn more about algebraic expression from brainly.com/question/11336599

#learnwithBrainly

5 0

3 years ago

Read 2 more answers

Find the equation of the line I in the following picture. The shapes under the lines are squares

julia-pushkina [17]

If you look carefully at the graph, you may see that the slope of the line is
3-4 -1
---------------- = ------ = m
7-4 3

thus, you have the slope of the line and two points on the line. Suppose we
choose the point (4,4) and subst. the known slope and the coordinates of this point into the point-slope formula for the eqn of a str line:

y-y1 = m (x-x1)

y-4 = (-1/3)(x-4)

This is the desired equation. You could, if you wished, change this into slope-intercept form.

5 0

2 years ago