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2 years ago

15

I need help with length times widths times height

2 answers:

jeyben [28]2 years ago

4 0

Add all of the same numbers together then do the length times width times height. :)

Andreyy892 years ago

3 0

I think the answer is 210

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By showing your work how do you multiply 1728 by 405

Likurg_2 [28]

1728
x405
-------
8640
0000
6912
--------
699,840

6 0

2 years ago

Find the solution of this system of equations. Separate the x- and y- values with a comma. x-2y=-24 and x-y=4

defon

X - 2y = -24
x - y = 4

Isolate x in the first equation by adding 2y to both sides.
x = -24 + 2y

Now plug in this value of x into the second equation.
(-24 + 2y) - y = 4
Solve. Combine all like terms, 2y - y.
-24 + y = 4
Add 24 to both sides to isolate y.
y = 28

Now plug y back into the first equation to find x.
x - 2(28) = -24
x - 56 = -24
Add 56 to both sides to isolate x.
x = 32

The solution is (32, 28).

6 0

2 years ago

Select the answer choice that is equivalent to the expression given belon.

uranmaximum [27]

Answer:

$23x - 10$

Step-by-step explanation:

To the find the equivalent of $9(2x - 3) + 5x + 17$ , evaluate the expression. Start by opening the bracket.

$9*2x - 9*3 + 5x + 17$

$18x - 27 + 5x + 17$

Pair like terms

$18x + 5x - 27 + 17$

$23x - 10$

The equivalent of $9(2x - 3) + 5x + 17$ is $23x - 10$

6 0

3 years ago

Describe the following sequence as arithmetic, geometric or neither. 2,4,8,12,14...

natka813 [3]

It should be neither. There is no pattern

6 0

2 years ago

Read 2 more answers

In a parallelogram ABCD point K belongs to diagonal BD so that BK:DK=1:4. If the extension of AK meets BC at point E, what is th

olya-2409 [2.1K]

Answer:

$\frac{BE}{EC} =\frac{1}{3}$

Step-by-step explanation:

In the diagram below we have

ABCD is a parallelogram. K is the point on diagonal BD, such that

$\frac{BK}{CK} =\frac{1}{4}$

And AK meets BC at E

now in Δ AKD and Δ BKE

∠AKD =∠BKE ( vertically opposite angles are equal)

since BC ║ AD and BD is transversal

∠ADK = ∠KBE ( alternate interior angles are equal )

By angle angle (AA) similarity theorem

Δ ADK and Δ EBK are similar

so we have

$\frac{AD}{BE} =\frac{DK}{BK}$

$\frac{AD}{BE} =\frac{4}{1}$

$\frac{BC}{BE}=\frac{4}{1}$ ( ABCD is parallelogram so AD=BC)

$\frac{BE+EC}{BE}=\frac{4}{1}$ ( BC= BE+EC)

$\frac{BE}{BE} +\frac{EC}{BE}=\frac{4}{1}$

$1+\frac{EC}{BE}=4$

$\frac{EC}{BE}=3$ ( subtracting 1 from both side )

$\frac{EC}{BE}=\frac{3}{1}$

taking reciprocal both side

$\frac{BE}{EC} =\frac{1}{3}$

8 0

3 years ago