If Jeff hikes 1/2 a distance in 1/4 an hour how fast can he walk in 1/2 an hour?<br>

Maya's history test had 40 questions. She answered 95% of the items correctly. How many questions did Maya answer correctly?

ankoles [38]

Maya got 38 of them right

5 0

2 years ago

Read 2 more answers

compute the projection of → a onto → b and the vector component of → a orthogonal to → b . give exact answers.

Nina [5.8K]

$\text { Saclar projection } \frac{1}{\sqrt{3}} \text { and Vector projection } \frac{1}{3}(\hat{i}+\hat{j}+\hat{k})$

We have been given two vectors $\vec{a}$ and $\vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

we have $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$

The scalar projection of $\vec{b}$ onto $\vec{a}$ means the magnitude of the resolved component of $\vec{b}$ the direction of $\vec{a}$ and is given by

The scalar projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|}$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

The vector projection of $\vec{b}$ onto

$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

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3 0

11 months ago

Please HELP ME ON THIS MATH WRITTEN RESPONSE QUESTION.. THANK YOU

sergij07 [2.7K]

Answer:

Quadrilateral ABCD is not a square. The product of slopes of its diagonals is not -1.

Step-by-step explanation:

Point A is (-4,6)

Point B is (-12,-12)

Point C is (6,-18)

Point D is (13,-1)

Given that the diagonals of a square are perpendicular to each other;

We know that the product of slopes of two perpendicular lines is -1.

So, slope(m) of AC × slope(m) of BD should be equal to -1.

Slope of AC = (Change in y-axis) ÷ (Change in x-axis) = (-18 - 6) ÷ (6 - -4) = -24/10 = -2.4

Slope of BD = (Change in y-axis) ÷ (Change in x-axis) = (-1 - -12) ÷ (13 - -12) = 11/25 = 0.44

The product of slope of AC and slope of BD = -2.4 × 0.44 = -1.056

Since the product of slope of AC and slope of BD is not -1 hence AC is not perpendicular to BD thus quadrilateral ABCD is not a square.

4 0

3 years ago

How can you tell whether a positive or negative association exists

stira [4]

Answer:

You can tell whether or not something is positive or negative by looking at the specific number for example if you look at this number -5 you know its a negative number because it has a - sign.Although when its positive it does not have any kind of format of signs according to my calculations.

Hope this helps!

Please give me 5 stars if it did i would very much appreciate it !!

6 0

2 years ago

Find the Missing Numerator and Denominator

n200080 [17]

Answer:

missing denominator: 5

missing numerator: 3

Step-by-step explanation:

You know the product of fractions is formed by multiplying numerators and multiplying denominators.

$\dfrac{3}{d}\times\dfrac{n}{4}=\dfrac{3n}{4d}=\dfrac{9}{20}$

This gives rise to two equations:

3n = 9 ⇒ n = 3

4d = 20 ⇒ d = 5

The missing denominator is 5; the missing numerator is 3.

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$\dfrac{3}{5}\times\dfrac{3}{4}=\dfrac{9}{20}$

7 0

2 years ago

If Jeff hikes 1/2 a distance in 1/4 an hour how fast can he walk in 1/2 an hour?​

If Jeff hikes 1/2 a distance in 1/4 an hour how fast can he walk in 1/2 an hour?