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3 years ago

(6) Find a and b as well as the ratio c:d in the figure below.

Mathematics

1 answer:

polet [3.4K]3 years ago

6 0

Answer:

i have a Similar question I’m stuck on, just remember z rule, vertically opposite etc.

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Can someone help me D: thanks in advance :D

max2010maxim [7]

Answer:

3) 27 cubed

4) 72 cubed

5) 125 cubed

8) 27 cubed. 3 is the length, 3 is the width, and 3 is the height.

9) 72 cubed. 6 is the length, 3 is the width, and 4 is the height.

10) 125 cubed. 5 is the length, 5 is the width, and 5 is the height.

Step-by-step explanation

Here's what to show if your teacher requires for you to show your work.

3) 3 x 3 x 3 = 27

4) 6 x 3 x 4 = 72

5) 5 x 5 x 5 = 125

8) 3 x 3 x 3 = 27

9) 6 x 3 x 4 = 72

10) 5 x 5 x 5 = 125

These are the ones I'd suggest putting the up and down form on.

10) and 5) Also put 5 x 5 = 25. 4) and 9) Also put 6 x 3 = 18.

2 3

25 18

x 5 x 4

125 72

Hoped this answered everything! Feel free to ask me if there's something I missed! :)

4 0

2 years ago

What does "As Much" mean in math? I am doing homework and I don't want to fail so I really need help.

igor_vitrenko [27]

It means the same used in a question like this, is 25 as much as 25.

4 0

3 years ago

If RT = 6 and RS = 9, then RX = A - sqrt(54) B - 13.5 C - 7.2 D - 4

wolverine [178]

Answer:

Option D RX=4 units

Step-by-step explanation:

we know that

In the right triangle RTS

The cosine of angle TRS is equal to

cos(TRS)=RT/RS

substitute

cos(TRS)=6/9 -----> equation A

In the right triangle RTX

The cosine of angle TRX is equal to

cos(TRX)=RX/RT

substitute

cos(TRX)=RX/6 -----> equation B

∠TRS=∠TRX -----> is the same angle

Match equation A and equation B

6/9=RX/6

RX=6*6/9=4 units

5 0

3 years ago

∫c x sin y ds, C is the line segment from (0, 1) to (3, 5)

Mnenie [13.5K]

Parameterize the line segment $C$ by

$\mathbf r(t)=\langle x(t),y(t)\rangle=(1-t)\langle0,1\rangle+t\langle3,5\rangle=\langle3t,1+4t\rangle$

where $t\in[0,1]$ . Then

$\mathrm ds=\|\mathbf r'(t)\|\,\mathrm dt$
$\mathrm ds=\|\langle3,4\rangle\|\,\mathrm dt$
$\mathrm ds=5\,\mathrm dt$

So the integral is

$\displaystyle\int_Cx\sin y\,\mathrm ds=5\int_0^1x(t)\sin y(t)\,\mathrm dt$
$=\displaystyle5\int_0^13t\sin(1+4t)\,\mathrm dt$
$=\dfrac{15}{16}(\sin5-\sin-4\cos5)\approx-2.7516$

3 0

3 years ago

The point (5,4) lies on a circle. what is the length of the radius of this circle if the center is located at (3,2)

Art [367]

Use the x and y values from your point and the h and k from the center to write the equation you need to solve for r. The general form of the equation for a circle is $(x-h) ^{2} +(y-k) ^{2} = r^{2}$ . We will fill in accordingly: $(5-3) ^{2}+(4-2) ^{2} = r^{2}$ . Solve for r to get that $r= \sqrt{8}$

4 0

3 years ago