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1 year ago

Here is a hanger: Write an equation to represent the hanger: __________________

Mathematics

2 answers:

Elenna [48]1 year ago

7 0

Answer:

hanger length X hanger height X hanger width

Step-by-step explanation:

MArishka [77]1 year ago

7 0

Answer:

3w = 25

Step-by-step explanation:

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Use the drop-down menus to complete the proof. By the unique line postulate, you can draw only one segment, AC BC CD. Using the

inn [45]

The first one is; BC
The second one is; Reflection
The third one is; A
The fourth one is; Length

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2 years ago

Read 2 more answers

Triangle J K L is shown. Angle J K L is a right angle. An altitude is drawn from point K to point M on side L J to form a right

kompoz [17]

Answer:

$LJ=15\ units$

Step-by-step explanation:

see the attached figure to better understand the problem

step 1

Find the length side KJ

In the right triangle JKM

Applying the Pythagoras Theorem

$KJ^{2}=JM^{2}+KM^{2}$

we have

$JM=3\ units$

$KM=6\ units$

substitute

$KJ^{2}=3^{2}+6^{2}$

$KJ^{2}=45}$

$KJ=\sqrt{45}\ units$

simplify

$KJ=3\sqrt{5}\ units$

step 2

Find the value of cosine of angle MJK in the right triangle JKM

$cos(JKM)=JM/KJ$

substitute the values

$cos(JKM)=\frac{3}{3\sqrt{5}}$

simplify

$cos(JKM)=\frac{\sqrt{5}}{5}$ -----> equation A

step 3

Find the value of cosine of angle MJK in the right triangle JKL

$cos(JKM)=KJ/LJ$

we have

$KJ=3\sqrt{5}\ units$

$cos(JKM)=\frac{\sqrt{5}}{5}$ ----> remember equation A

substitute the values

$\frac{\sqrt{5}}{5}=\frac{3\sqrt{5}}{LJ}$

Simplify

$LJ=5(3)=15\ units$

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2 years ago

Read 2 more answers

The time that a butterfly lives after emerging from its chrysalis can be modelled by a random variable T, the model here taking

olasank [31]

Xd answer is d I hope u find the answer

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2 years ago

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.

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2 years ago

Find the percent of change. Round to the nearest tenth, if necessary.

S_A_V [24]

Answer:

20.487%

Step-by-step explanation:

divide the second number by the first. then do one minus that number

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2 years ago