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4 years ago

Rewrite the equation by completing the square. x^2 - 14x + 40 = 0

Mathematics

1 answer:

Angelina_Jolie [31]4 years ago

7 0

Answer:

Solve the equation for

x^2 - 14x+49

(X - 7 )^2

So x= 10,4

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Find the nth term of this quadratic sequence 4, 7, 12, 19, 28,...

Len [333]

Tn=n²+3

from 4 to 7....the difference is 3, then from 7 to 12 the difference is 5 and from 12 to 19 the difference is 7

therefore we constantly add 2 to the difference

4 0

3 years ago

Write the equation of the line that passes through the points (5, -1) and

trapecia [35]

Answer:

y = -1

Step-by-step explanation:

First, you find the slope by using the slope formula and get 0. Now, choose one coordinate and plug it into the equation y = mx + b along with the slope. If you chose the first coordinate, the equation would be -1 = (0)(5) + b. After solving for b (the y-intercept) you would get -1. Lastly, you rewrite the equation as y = -1.

7 0

3 years ago

How many feet are there in 24 yards? A. 288 B. 8 C. 2 D. 72

Mkey [24]

First of all we need to know that 1 yard has 3 feet inside of it
but in order to find 24 yards we would have to multiply by 3 sooo
24 x 3 = 72
so that means there are 72 feet in 24 yards.
In other words the answer is D:)
hope this helps:P
MARK ME BRAINLIEST!

6 0

4 years ago

Read 2 more answers

Evaluate the integral, show all steps please!

Aloiza [94]

Answer:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x=\dfrac{x}{9\sqrt{9-x^2}} +\text{C}$

Step-by-step explanation:

Fundamental Theorem of Calculus

$\displaystyle \int \text{f}(x)\:\text{d}x=\text{F}(x)+\text{C} \iff \text{f}(x)=\dfrac{\text{d}}{\text{d}x}(\text{F}(x))$

If differentiating takes you from one function to another, then integrating the second function will take you back to the first with a constant of integration.

Given indefinite integral:

$\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x$

Rewrite 9 as 3² and rewrite the 3/2 exponent as square root to the power of 3:

$\implies \displaystyle \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x$

Integration by substitution

$\boxed{\textsf{For }\sqrt{a^2-x^2} \textsf{ use the substitution }x=a \sin \theta}$

$\textsf{Let }x=3 \sin \theta$

$\begin{aligned}\implies \sqrt{3^2-x^2} & =\sqrt{3^2-(3 \sin \theta)^2}\\ & = \sqrt{9-9 \sin^2 \theta}\\ & = \sqrt{9(1-\sin^2 \theta)}\\ & = \sqrt{9 \cos^2 \theta}\\ & = 3 \cos \theta\end{aligned}$

Find the derivative of x and rewrite it so that dx is on its own:

$\implies \dfrac{\text{d}x}{\text{d}\theta}=3 \cos \theta$

$\implies \text{d}x=3 \cos \theta\:\:\text{d}\theta$

Substitute everything into the original integral:

$\begin{aligned}\displaystyle \int \dfrac{1}{(9-x^2)^{\frac{3}{2}}}\:\:\text{d}x & = \int \dfrac{1}{\left(\sqrt{3^2-x^2}\right)^3}\:\:\text{d}x\\\\& = \int \dfrac{1}{\left(3 \cos \theta\right)^3}\:\:3 \cos \theta\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{\left(3 \cos \theta\right)^2}\:\:\text{d}\theta \\\\ & = \int \dfrac{1}{9 \cos^2 \theta} \:\: \text{d}\theta\end{aligned}$

Take out the constant:

$\implies \displaystyle \dfrac{1}{9} \int \dfrac{1}{\cos^2 \theta}\:\:\text{d}\theta$

$\textsf{Use the trigonometric identity}: \quad\sec^2 \theta=\dfrac{1}{\cos^2 \theta}$

$\implies \displaystyle \dfrac{1}{9} \int \sec^2 \theta\:\:\text{d}\theta$

$\boxed{\begin{minipage}{5 cm}\underline{Integrating $\sec^2 kx$}\\\\$\displaystyle \int \sec^2 kx\:\text{d}x=\dfrac{1}{k} \tan kx\:\:(+\text{C})$\end{minipage}}$