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3 years ago

Which equations represents a line that passes through (4, 1/3) and has a slope of 3/4

Mathematics

2 answers:

suter [353]3 years ago

8 0

Answer:

y=3/4x-8/3

Step-by-step explanation:

y-y1=m(x-x1)

y-1/3=3/4(x-4)

y-1/3=3/4x-12/4

y-1/3=3/4x-3

y=3/4x-3+1/3

y=3/4x-9/3+1/3

y=3/4x-8/3

docker41 [41]3 years ago

6 0

Answer:

y = 3/4x - 8/3

Step-by-step explanation:

y = 3/4x + b

1/3 = 3/4(4) + b

1/3 = 3 + b

-8/3 = b

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Solve for x , (2x-1)=5(x+1) no links

kobusy [5.1K]

Answer:

x = -2

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

2 times 10 to the 4th subtracted to 7 to the 3ed

alekssr [168]

2 x 10 4th power - 7 3rd power

2 x 10,000 - 343

20,000 - 343

19,657

I think this is the answer to your problem. Hope this helped!

-TTL

6 0

3 years ago

Which of the expressions below can be factored using the difference of squares method?

Dmitry [639]

Answer: A. $9x^2 - 16y^2$ .

Step-by-step explanation:

A difference of squares problem is factored as : $a^2 - b^2 = (a + b)(a - b)$
For this both terms in the form of squares and a minus sign between them.

From all the options only option A has both terms square and they minus sign between them.

Also, $9x^2 - 16y^2= (3x)^2-(4y)^2=(3x+4y)(3x-4y)$

Here a= 3x and b= 4

Hence, the correct answer is A. $9x^2 - 16y^2$ .

3 0

3 years ago

The centers for disease control and prevention reported that 25% of baby boys 6-8 months old in the united states weigh more tha

melisa1 [442]

Answer:

0.1971 ( approx )

Step-by-step explanation:

Let X represents the event of weighing more than 20 pounds,

Since, the binomial distribution formula is,

$P(x)=^nC_r p^r q^{n-r}$

Where, $^nC_r=\frac{n!}{r!(n-r)!}$

Given,

The probability of weighing more than 20 pounds, p = 25% = 0.25,

⇒ The probability of not weighing more than 20 pounds, q = 1-p = 0.75

Total number of samples, n = 16,

Hence, the probability that fewer than 3 weigh more than 20 pounds,

$P(X$

$=^{16}C_0 (0.25)^0 (0.75)^{16-0}+^{16}C_1 (0.25)^1 (0.75)^{16-1}+^{16}C_2 (0.25)^2 (0.75)^{16-2}$

$=(0.75)^{16}+16(0.25)(0.75)^{15}+120(0.25)^2(0.75)^{14}$

$=0.1971110499$

$\approx 0.1971$

3 0

3 years ago

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})$\end{minipage}}$