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3 years ago

Please i need help with this question (7^6)^2

Mathematics

1 answer:

elena-14-01-66 [18.8K]3 years ago

4 0

The answer is 7^12 but you can still simplify it

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2. Kendall wants to buy 8 cans of cat : 8.

GREYUIT [131]

Answer: $11.44

Step-by-step explanation:

I divided 4.29 by 3 which was $1.43 then i multiplied it by 8 to get $11.44

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3 years ago

Read 2 more answers

2. How many milliliters are in 1 pint? A. 120 B. 240 C. 480 D. 960

svet-max [94.6K]

Answer: C- 480

Step-by-step explanation:

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3 years ago

Slope between (-3,-3) and (-5,-2)

Katyanochek1 [597]

Answer:

idontquiteknowbut

Step-by-step explanation:

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4 years ago

The operation manager at a tire manufacturing company believes that the mean mileage of a tire is 48,637 miles, with a variance

Brut [27]

Answer:

$P(x < -778) = 0$

Step-by-step explanation:

Given

$\bar x = 48673$

$\sigma^2 = 11282880$

$n = 143$

Required

$P(x$

First, we calculate the z score

$z = \frac{x}{\sqrt{\sigma^2}/n}$

So, we have:

$z = \frac{-778}{\sqrt{11282880}/143}$

$z = \frac{-778}{3359.0/143}$

$z = \frac{-778}{23.49}$

$z = -33.12$

So:

$P(x < -778) = P(z < -33.12)$

From z score probability, we have:

$P(x < -778) = 0$

4 0

3 years ago

g find the 2 components of vector b = 2i + j - 3k, one parallel to a = 3i - j and another one perpendicular to a

nika2105 [10]

Answer:

The components of $\vec{b}$ parallel and perpendicular to $\vec {a}$ are $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ and $\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$ , respectively.

Step-by-step explanation:

Let be $\vec b = 2\,i+j-3\,k$ and $\vec a = 3\,i-j$ , the component of $\vec b$ parallel to $\vec a$ is calculated by the following expression:

$\vec b_{\parallel} = (\vec b \bullet \hat{a}) \cdot \hat{a}$

Where $\hat{a}$ is the unit vector of $\vec a$ , dimensionless and $\bullet$ is the operator of scalar product.

The unit vector of $\vec a$ is:

$\hat{a} = \frac{\vec {a}}{\|\vec a\|}$

Where $\|\vec {a}\|$ is the norm of $\vec a$ , whose value is determined by Pythagorean Theorem.

The component of $\vec{b}$ parallel to $\vec {a}$ is:

$\|\vec {a}\| = \sqrt{3^{2}+(-1)^{2}+0^{2}}$

$\|\vec {a}\| = \sqrt{10}$

$\hat{a} = \frac{1}{\sqrt{10}} \cdot (3\,i-j)$

$\hat{a} = \frac{3}{\sqrt{10}}\,i -\frac{1}{\sqrt{10}} \,j$

$\vec{b}\bullet \hat{a} = (2)\cdot \left(\frac{3}{\sqrt{10}} \right)+(1)\cdot \left(-\frac{1}{\sqrt{10}} \right)+(-3)\cdot \left(0\right)$

$\vec b \bullet \hat{a} = \frac{5}{\sqrt{10}}$

$\vec b_{\parallel} = \frac{5}{\sqrt{10}}\cdot \left(\frac{3}{\sqrt{10}}\,i-\frac{1}{\sqrt{10}}\,j \right)$

$\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$

Now, the component of $\vec {b}$ perpendicular to $\vec{a}$ is found by vector subtraction:

$\vec{b}_{\perp} = \vec {b}-\vec {b}_{\parallel}$

If $\vec b = 2\,i+j-3\,k$ and $\vec {b}_{\parallel} = \frac{3}{2}\,i-\frac{1}{2}\,j$ , then:

$\vec{b}_{\perp} = (2\,i+j-3\,k)-\left(\frac{3}{2}\,i-\frac{1}{2}\,j \right)$

$\vec b _{\perp} = \frac{1}{2}\,i+\frac{3}{2}\,j-3\,k$

4 0

3 years ago