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Mazyrski [523]
3 years ago
12

Function 1 is defined by the following table.

Mathematics
1 answer:
Ksju [112]3 years ago
8 0
Function 1 I think because it’s has the greater slope
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A plane is flying at 31,348 feet. It needs to rise to 36,000 feet in two stages. In stage 1, it rises 5% of its initial altitude
In-s [12.5K]

Answer:

well, 5% of 31348 is 1567.4

That leaves her 3084.6 feet to go.

And as you know, time = distance/speed

Step-by-step explanation:

6 0
3 years ago
What reasoning and explanations can be used when solving radical equations and how do extraneous solutions arise from radical eq
Mkey [24]

Answer:

We know that, a 'radical equation' is an equation that contains radical expressions, which further are the expressions containing radicals ( square roots and other roots of numbers ).

In order to solve radical equations, we use the rules of exponents and basic algebraic properties.

The common reasoning to use while solving a radical equation is:

1. Isolate the radical expression.

2. Square both sides of the equation to remove radical.

3. After removing the radical, solve the equation to find the unkown variable

4. Check the answer for the errors occurred by removing the radicals.

For e.g. \sqrt{x}-3=5

i.e. \sqrt{x}-3+3=5+3 ( Adding 3 on both sides )

i.e. \sqrt{x}=8

i.e. (\sqrt{x})^{2} =8^{2}

i.e. x=64.

So, the solution the the radical equation \sqrt{x}-3=5 is x = 64.

Further, we know that an 'extraneous solution' is that root of the radical equation which is not a root of the original equation and is excluded from the domain.

for e.g. Take \sqrt{x+4} =x-2

i.e. (\sqrt{x+4})^{2} =(x-2)^2

i.e. x+4=x^2+4-4x

i.e. 0=x^2-5x

i.e. 0=x(x-5)

i.e. x = 0 and x = 5.

Substituting x = 0 in \sqrt{x+4} =x-2, gives \sqrt{0+4} =0-2 i.e. \sqrt{4} =-2 i.e. 2=-2, which is not possible.

So, x = 0 is a solution that does not satisfy the equation.

Hence, x = 0 is an extraneous solution.

7 0
4 years ago
Calculus help! Using the mean value theorem!
marta [7]
f is differentiable across its domain, so the MVT says there is some value of c in the open interval (a,b) such that

f'(c)=\dfrac{f(b)-f(a)}{b-a}

You have

f(x)=Ax^2+Bx+C\implies f'(x)=2Ax+B

so the equation above becomes

2Ac+B=\dfrac{(Ab^2+Bb+C)-(Aa^2+Ba+C)}{b-a}

Solve for c.

2Ac+B=\dfrac{A(b^2-a^2)+B(b-a)}{b-a}
2Ac+B=A(b+a)+B
2Ac=A(b+a)
c=\dfrac{b+a}2

so c is indeed the average of the endpoints, i.e. the midpoint.
8 0
3 years ago
Simplify the expression:<br> 10q–3q+2q2+3
saw5 [17]
If the 2q is raised to the second power the answer is 2q^2+7q+3
4 0
3 years ago
Read 2 more answers
H(x) = 9x2 + 1<br> PLSSS HELP
nataly862011 [7]

Answer: 82

Step-by-step explanation:

H=9(-3)^2+1

Simplify both sides of the equation.

h=81+1

81+1 =82

h=82

4 0
2 years ago
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