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3 years ago

Spike weighs twice as Snowball. How much does Spike weigh ?

Mathematics

1 answer:

kaheart [24]3 years ago

5 0

Answer:

spike weighs 18lbs

Step-by-step explanation:

doubled means multiply by 2 so 2*9=18

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GIVING BRANLIEST<br> 11t - 4t when t = 2

Artyom0805 [142]

Answer:

PLEASE GIVE BRAINLIEST.

Step-by-step explanation:

22 - 8 = 14

6 0

3 years ago

Read 2 more answers

What is the explicit rule for the geometric sequence 1/64, 1/16, 1/4, 1

deff fn [24]

Answer:

$\frac{1}{\sqrt[n]{64}}$

Step-by-step explanation:

$\frac{1}{\sqrt[1]{64}} =\frac{1}{64}$

$\frac{1}{\sqrt[2]{64}} =\frac{1}{16}$

$\frac{1}{\sqrt[3]{64}} =\frac{1}{4}$

$\frac{1}{\sqrt[4]{64}} =\frac{1}{1}$

5 0

3 years ago

Someone please help meeeeeeeee

Leya [2.2K]

Answer:

180-(90+28) =62

Step-by-step explanation:

The whole angle is 180 minus the 90° and 28°

6 0

3 years ago

Find the volume of the given solid. Bounded by the cylinder y2 + z2 = 16 and the planes x = 2y, x = 0, z = 0 in the first octant

erastovalidia [21]

Answer:

$\mathbf{ \dfrac{128}{3}}$

Step-by-step explanation:

Given that:

$y^2 + z^2 = 16; \\ \\ where; \ x = 2y , x =0, z= 0$

$Replace \ z = 0 \ \text{in the given equation ;} y^2+z^2=16$

$Then;$

$y^2 = 16 \\ \\ y = \sqrt{16} \\ \\ y = \pm 4$

$\text{So; in 1st octant \ limits of y }= 0 \to 4 \ \text{and for x }= 0 \to 2y$

∴

$\text{Volume of solid: (V) = }\int ^{4}_{y=0} \int ^{2y}_{x=0} \sqrt{16-y^2} \ dxdy \\ \\ = \int^4_{y=0} \sqrt{16 -y^2} \ (x)^{2y} _o \ dy \\ \\ = \int^4_{y=0}\sqrt{16 -y^2} (2y -0) \ dy \\ \\ V = \int^{4}_{y=0} \ 2y \sqrt{16 - y^2} \ dy$

$Now, let:\\\\ 16 - y^2 = u \\ \\ -2ydy = du \\ \\ 2ydy = -du \\ \\ Hence, when \ y = 0; u = 16 \\ \\ and \\ \\ when \ y = 4 \ then \ u = 0$

∴

$V = \int ^4 _{y=0} \ 2y \sqrt{16 -y^2 } \ dy \\ \\ = \int ^0_{16} - \sqrt{u}\ du \\ \\ = - \int^0_{16} \ u^{^{\dfrac{1}{2}}} \ du \\ \\ = \Bigg [\dfrac{u^{^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1 } \Bigg]^0_{16} \\ \\ =\Bigg[ -\dfrac{u^{\dfrac{3}{2}}}{\dfrac{3}{2}}^0_{16} \Bigg] \\ \\$

$= - \dfrac{2}{3}\Big( u^{\dfrac{3}{2}}\Big)^0_{16} \\ \\ = - \dfrac{2}{3} \Big(0^{^\dfrac{3}{2}}}- 16^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(0 - (4^2)^{^{\dfrac{3}{2}}}\Big ) \\ \\ = - \dfrac{2}{3} \Big(-(4)^{3}}\Big )$