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2 years ago

HELP ME OUT PLEASE!!

Mathematics

2 answers:

Zepler [3.9K]2 years ago

8 0

Answer:

may sagot na pala ok bye thanks sa pts

zhannawk [14.2K]2 years ago

5 0

Answer:

30!

Step-by-step explanation:

sorry if its wrong but hoped it helped :D

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In a candy mix there are 4 green bars for every 3 red bars. How many red bars are there if there are 200 green bars?

UkoKoshka [18]

150 red bars

4/3=200/150

3 0

3 years ago

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What statement is true about f(x)=2x

inna [77]

Answer:

Subtract

from both sides of the equation.

(

)

−

Step-by-step explanation:

5 0

3 years ago

Find the cost of painting a cubical block of side length 3.2 cm, if painting costs $5 per cm2.

SashulF [63]

<h3><u>Answer</u>:- </h3>

$\longrightarrow \sf \$ \: 307.2$

<h3><u>Solution:-</u></h3>

Since painting is done on outer surface of cubical block we will have to find its surface area which is given by-

$\green{ \underline { \boxed{ \sf{Surface \: area \:of \:Cube = 6a^2}}}}$

<u>where a is side of the cube </u>

$\begin{gathered}\\\quad \sf Surface \: area \:of \:Cube = 6\times(3.2)^2 \\\end{gathered}$

$\begin{gathered}\\\implies \quad \sf 6\times 3.2\times3.2 \\\end{gathered}$

$\begin{gathered}\\\implies \quad \sf 61.44 cm^2 \\\end{gathered}$

Now,

$\maltese \sf Cost \:of \:painting \: 1 \: cm² = \$ 5$

$\sf Cost \: of \: painting \: 61.44 \:cm²=61.44×5$

$\sf\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad=\$ 307.2$

$\leadsto$ Thus it would cost $ 307.2 to paint the entire cubical block.

3 0

2 years ago

SOMEONE PLEASE HELP ME ASAP WITH THIS AND SHOW WORK

Phantasy [73]

I’m sorry I can’t help but pray

5 0

3 years ago

A ball is launched straight up in the air from a height of 6 feet. Its velocity (feet/second) t seconds after launch is given by

german

Step-by-step explanation:

A ball is launched straight up in the air from a height of 6 feet. The velocity as a function of time t is given by :

f(t) = -32 t+285

Height of the ball is :

$h(t)=\int\limits{f(t){\cdot} dt}\\\\h(t)=\int\limits{(-32t+285){\cdot} dt}\\\\h(t)=-16t^2+285t+C$

C is constant. Here the ball is launched from a height of 6 feet. So,

$h(t)=-16t^2+285t+6$

At t = 2 s, $h(t)=-16(2)^2+285(2)+6=512\ m$

At t = 9 s, $h(t)=-16(9)^2+285(9)+6=1275\ m$

Between 2 s and 9 s, the ball's height changed is : 1275 - 512 = 763 m.

6 0

3 years ago