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2 years ago

What is the volume of the triangular prism in the picture? (sorry if the picture is blurry)

Mathematics

2 answers:

bixtya [17]2 years ago

6 0

392. Because 8*7*14= 784
(784 times .5 is 392

Leviafan [203]2 years ago

3 0

Can I tell u the answer

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I NEED HELP PLEASE!!!!!!!!!!!!!!!!!!!!!!!!!

diamong [38]

Geez that is real complicated sorry I’m not sure just go with your gut

6 0

3 years ago

9. Cal is buying T-shirts and shorts. T-shirts

damaskus [11]

Hey there!

Step By Step:

Well so we know that 12 is about half of an 25. So what we are gonna do is to buy 3 of 12$ and 2 of 25$.

3x12=36 and 2x25=50.

With adding those two last numbers, you’ll get $86.

Graph would be down below.

Hope it helps :D

3 0

3 years ago

Find the value of the expressions 3x^3-2y^3-6x^2y^2+xy for x=2/3 and y=1/2

fiasKO [112]

Hi!

$3x^3-2y^3-6x^2y^2+xy=\\\\=3\cdot(\frac{2}{3})^3-2\cdot(\frac{1}{2})^3-6\cdot(\frac{2}{3})^2\cdot(\frac{1}{2})^2+\frac{2}{3}\cdot\frac{1}{2}=\\\\=3\cdot\frac{8}{27}-2\cdot\frac{1}{8}-6\cdot\frac{4}{9}\cdot\frac{1}{4}+\frac{1}{3}=\\\\=\frac{8}{9}-\frac{1}{4}-\frac{2}{3}+\frac{1}{3}=\frac{8}{9}-\frac{1}{4}-\frac{1}{3}=\frac{32}{36}-\frac{9}{36}-\frac{12}{36}=\boxed{\frac{11}{36}}$

5 0

2 years ago

Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y

irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

$y'(x)= x^2y(x)-\frac12y^2(x)$

Putting the value of y'(x) in equation (1)

$y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))$

Substituting x =0 and h= 0.2

$y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]$

$\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]$ [∵ y(0) =3 ]

$\Rightarrow y(0.2)\approx 2.7$

Substituting x =0.2 and h= 0.2

$y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]$

$\Rightarrow y(0.4)\approx 2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]$

$\Rightarrow y(0.4)\approx 1.9926$

Substituting x =0.4 and h= 0.2

$y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]$

$\Rightarrow y(0.6)\approx 1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]$

$\Rightarrow y(0.6)\approx 1.6593$

Substituting x =0.6 and h= 0.2

$y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]$

$\Rightarrow y(0.8)\approx 1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]$

$\Rightarrow y(0.6)\approx 0.8800$

Substituting x =0.8 and h= 0.2

$y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]$

$\Rightarrow y(1.0)\approx 0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]$

$\Rightarrow y(1.0)\approx 0.9152$

Therefore the value of y(1)= 0.9152.

4 0

3 years ago

A square with one side length represented by an expression is shown below

crimeas [40]

It is mentioned in the problem that the shape is a square, hence, we could say that all side's measurement is the same and it is equal. It can be written in a different arrangement of terms, but still, the value to be solved is the same. The sides can be written in the following forms:
S1=6(3X+8)+32+12X
S2=18X+48+32+12X
S3=18X+80+12X
S4=30X+80

3 0

3 years ago