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3 years ago

Help asap please and thank you. true or false?

Mathematics

2 answers:

Mkey [24]3 years ago

8 0

True! Im pretty sure!

san4es73 [151]3 years ago

6 0

The answer is true
:)

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40% of one spig is a spoog. 25% of a speeg is a spoog. 70% of a speeg is a spug. What percent of 1 spig is 5 spugs

adoni [48]

Answer:

20% of a spig is a spug.

Step-by-step explanation:

Spig = I

Spoog = O

Speeg = E

Spug = U

.4I = O

.25E = O

.7E = U

1I = 5 U To get the percentage of spigs we divideeach side by 5 to get spugs.

1 I / 5 = 5 U / 5

1/5 I = U

1/5 = .2 = 20%

.2I = U

20% of a spig is a spug.

7 0

3 years ago

Your new phew is standing on his deck which is 4 feet off the ground. He tosses his toy up into the air. The equation h(t) = -2t

Tpy6a [65]

Answer:

The height of the toy is 9 feet after 1 second

Step-by-step explanation:

we have

$h(t)=-2t^2+7t+4$

where

h(t) is the toys height from the ground

t is the time in seconds

For t=1 sec

substitute the value of t in the quadratic equation and solve for h(t)

$h(1)=-2(1)^2+7(1)+4$

$h(1)=9\ ft$

3 0

4 years ago

Hey y’all pls help me , what’s the answer :))

klio [65]

4th one. its simple math bro

7 0

3 years ago

Read 2 more answers

Write three different sentences that could describe the relationship between the quantities $27, $81

Olenka [21]

1) 27+x=81
2)27+81=108
3)81-x=27

8 0

3 years ago

Find a particular solution to y" - y + y = 2 sin(3x)

leonid [27]

Answer with explanation:

The given differential equation is

y" -y'+y=2 sin 3x------(1)

Let, y'=z

y"=z'

$\frac{dy}{dx}=z\\\\d y=zdx\\\\y=z x$

Substituting the value of , y, y' and y" in equation (1)

z'-z+zx=2 sin 3 x

z'+z(x-1)=2 sin 3 x-----------(1)

This is a type of linear differential equation.

Integrating factor

$=e^{\int (x-1) dx}\\\\=e^{\frac{x^2}{2}-x}$

Multiplying both sides of equation (1) by integrating factor and integrating we get

$\rightarrow z\times e^{\frac{x^2}{2}-x}=\int 2 sin 3 x \times e^{\frac{x^2}{2}-x} dx=I$

$I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx -\int \frac{2\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx-\frac{2I}{3}\\\\\frac{5I}{3}=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{3}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{3} dx\\\\I=\frac{-2\cos 3x e^{\fra{x^2}{2}-x}}{5}+\int\frac{2x\cos 3x e^{\fra{x^2}{2}-x}}{5} dx$

8 0

3 years ago