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3 years ago

Can I get help on this please

Mathematics

1 answer:

Misha Larkins [42]3 years ago

4 0

Answer:

5. You can round question a up to 100 + 55 (then do the counting)

5. You can round question b up to 245 - 100 (then do the counting)

You can round question c up to 100 x 5 (then do the counting)

You can round question d up to 100 x 20 (then subtract accordingly)

Please rank Brainliest if this helps!

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5c + cd when c 1/5 and d = 15

vichka [17]

5c + cd
=5(1/5) + (1/5)(15)
= 1 + 3
= 4

5 0

3 years ago

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HELP MEEEE PLZ. i need to Find the surface area and volume

lisabon 2012 [21]

Answer:

A ≈ 494.80 $mm^{2}$

V = 827.02 $mm^{3}$

Step-by-step explanation:

height ( h ) = 13 mm

radius ( r ) = $\frac{9mm}{2}$ = 4.5 mm

Surface area = $2\pi rh + 2\pi r$

= 2π(4.5)(13) + 2π(4.5)

≈ 494.80 $mm^{2}$

Volume = $\pi r^{2} h$

= $\pi (4.5)^{2} (13)$

= 827.02 $mm^{3}$

6 0

2 years ago

Will mark correct answer branliest!!!

WITCHER [35]

Answer: A = 4 P = π PS = -π/2

Step-by-step explanation:

y = A sin(Bx - C) + D where

A is the amplitude
Period = 2π/B
Phase Shift = C/B
D is the vertical shift

Given: y = -4 sin(2x + π) - 3

A (amplitude) = 4 the negative sign is a reflection, not amplitude
B = 2
C = -π
D = -3

Amplitude = 4

$\text{Period}=\dfrac{2\pi}{B}\quad =\quad \dfrac{2\pi}{2}\quad =\quad \pi$

$\text{Phase Shift}=\dfrac{C}{B}\quad =\quad \dfrac{-\pi}{2}\quad =\quad -\dfrac{\pi}{2}$

4 0

3 years ago

Find the integral, using techniques from this or the previous chapter. ∫x(8-x)3/2 dx

Soloha48 [4]

Answer:

$\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C$

Step-by-step explanation:

For this case we need to find the following integral:

$\int x(8-x)^{3/2}dx$

And for this case we can use the substitution $u = 8-x$ from here we see that $du = -dx$ , and if we solve for x we got $x = 8-u$ , so then we can rewrite the integral like this:

$\int x(8-x)^{3/2}dx= \int (8-u) u^{3/2} (-du)$

And if we distribute the exponents we have this:

$\int x(8-x)^{3/2}dx= - \int 8 u^{3/2} + \int u^{5/2} du$

Now we can do the integrals one by one:

$\int x(8-x)^{3/2}dx= -8 \frac{u^{5/2}}{\frac{5}{2}} + \frac{u^{7/2}}{\frac{7}{2}} +C$

And reordering the terms we have"

$\int x(8-x)^{3/2}dx= -\frac{16}{5} u^{\frac{5}{2}} +\frac{2}{7} u^{\frac{7}{2}} +C$

And rewriting in terms of x we got:

$\int x(8-x)^{3/2}dx= -\frac{16}{5} (8-x)^{\frac{5}{2}} +\frac{2}{7} (8-x)^{\frac{7}{2}} +C$

And that would be our final answer.

8 0

2 years ago

Can someone help me

Nikolay [14]

Omg this looks hard

4 0

2 years ago

Read 2 more answers