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3 years ago

Help with multi step inequalities please

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

5 0

Answer:

B. The solution is valid because all steps to solve the inequality for F are correct.

Step-by-step explanation:

F - 32 ≤ 0

Add 32 to both sides of the equation to have;

F -32 + 32 ≤ 0 + 32

F ≤ 0 + 32

F ≤ 32

It can be observed that to solve for F, the steps are correct. Thus the solution is valid. Therefore, the correct choice in the given question is the solution is valid because all steps to solve the inequality for F are correct.

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the sum of 49,000 naira is to be shared among A, B, and C in the ratio of 1:2:4, respectively. find how much each wilk receive.

Natalka [10]

Answer:

#7000, #14000 and #28000 respectively

Step-by-step explanation:

#49000 to be shared in 1:2:4

49000 ÷ (1 + 2 + 4) = 49000/7 = #7000

A will get (7000×1) = #7000

B will get (7000×2) = #14000

C will get (7000×4) = #28000

3 0

2 years ago

Solve for A Exact (no rounding, no calculator needed) 0 = A^2 + 13A + 36

marysya [2.9K]

Answer:

-4 or -9

Step-by-step explanation:

-a^2-13a-36=0

(-a-4)(a+9)=0

-a-4=0

a+9=0

a= -4, -9

3 0

2 years ago

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Marrrta [24]

Given:

$f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6$

Required:

$We\text{ need to find }F^{\prime}(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.$

Explanation:

Given equation is

$F(x)=\sqrt[4]{f(x)g(x)}.$ $F(x)=(f(x)g(x))^{\frac{1}{4}}$ $F(x)=f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}}$

Differentiate the given equation for x.

$Use\text{ }(uv)^{\prime}=uv^{\prime}+vu^{\prime}.\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.$

$F^{\prime}(x)=f(x)^{\frac{1}{4}}(\frac{1}{4}g(x)^{\frac{1}{4}-1})g^{\prime}(x)+g(x)^{\frac{1}{4}}(\frac{1}{4}f(x)^{\frac{1}{4}-1})f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1}{4}-\frac{1\times4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1}{1}-\frac{1\times4}{4}}f^{\prime}(x)$ $=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{1-4}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{1-4}{4}}f^{\prime}(x)$ $F^{\prime}(x)=\frac{1}{4}f(x)^{\frac{1}{4}}g(x)^{\frac{-3}{4}}g^{\prime}(x)+\frac{1}{4}g(x)^{\frac{1}{4}}f(x)^{\frac{-3}{4}}f^{\prime}(x)$

Replace x=3 in the equation.

$F^{\prime}(3)=\frac{1}{4}f(3)^{\frac{1}{4}}g(3)^{\frac{-3}{4}}g^{\prime}(3)+\frac{1}{4}g(3)^{\frac{1}{4}}f(3)^{\frac{-3}{4}}f^{\prime}(3)$ $Substitute\text{ }f(3)=8,f^{\prime}(3)=-4,g(3)=2,\text{ and }g^{\prime}(3)=-6\text{ in the equation.}$ $F^{\prime}(3)=\frac{1}{4}(8)^{\frac{1}{4}}(2)^{\frac{-3}{4}}(-6)+\frac{1}{4}(2)^{\frac{1}{4}}(8)^{\frac{-3}{4}}(-4)$ $F^{\prime}(3)=\frac{-6}{4}(8)^{\frac{1}{4}}(2^3)^{\frac{-1}{4}}+\frac{-4}{4}(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}(8)^{\frac{1}{4}}(8)^{\frac{-1}{4}}-(2)^{\frac{1}{4}}(8^3)^{\frac{-1}{4}}$ $F^{\prime}(3)=\frac{-3}{2}\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}$ $F^{\prime}(3)=\frac{-3}{2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-3\times2}{2\times2}-\frac{1}{4}$ $F^{\prime}(3)=\frac{-6-1}{4}$ $F^{\prime}(3)=\frac{-7}{4}$

Final answer:

$F^{\prime}(3)=\frac{-7}{4}$

8 0

10 months ago

Please answer and explanation

mrs_skeptik [129]

61.23 square inches of metal is needed to create a cylindrical can.

Solution:

Diameter of the base = 3 in

Radius of the base = 3 ÷ 2 = 1.5 in

Height of the cylinder = 5 in

The value of π = 3.14

To find the surface area of the cylinder:

Surface area of the cylinder = $2 \pi r^{2}+2 \pi r h$

= 2 × 3.14 × (1.5)² + 2 × 3.14 × 1.5 × 5

= 14.13 + 47.1

Surface area of the cylinder = 61.23 sq. in

Hence 61.23 square inches of metal is needed to create a cylindrical can.

3 0

3 years ago

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boyakko [2]

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2 years ago