A 9 in The Hundreds place and a six in the hundred Thousands place

Renae and her family are on vacation. During the drive from Denver to Rocky Mountain National Park, she saw 48 travel trailers a

Setler [38]

Answer:

27:48

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Can someone help me and explain? I will mark brainlest. ♡

Sedbober [7]

Answer:

$p'(4) = -3$

$q'(8) = \frac{1}{4}\\$

Step-by-step explanation:

For p'(4):

$p(x) = f(x)g(x) \\ p'(x) = \frac{d}{dx}(f(x)g(x)) \\ p'(x) = f'(x)g(x) +f(x)g'(x)$

$p'(4) = f'(4)g(4) + f(4)g'(4) \\ p'(4) = (-1)(3) +(7)(0) \\ p'(4) = -3$

For q'(8):

$q(x) = \frac{f(x)}{g(x)} \\ q'(x)= \frac{d}{dx}(\frac{f(x)}{g(x)}) \\ q'(x) = \frac{f'(x)g(x) -f(x)g'(x)}{{g(x)}^2}$

$q'(8) = \frac{f'(8)g(8) -f(8)g'(8)}{{g(8)}^2} \\ q'(8) = \frac{(2)(2) -(6)(\frac{1}{2})}{{2}^2} \\ q'(8) = \frac{4 -3}{4} \\ q'(8) = \frac{1}{4}$

4 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

#SPJ4

3 0

9 months ago

Find each missing length, make sure you give both exact AND approximate. Brainliest + 50pts

goldenfox [79]

#1

B²=15²-7²
B²=225-49
B²=176
B=13.26(exact)
B=13.3(approx)

#2

B²=17²-7²
B²=289-49
B²=240
B=15.49(exact)
B=15.5

#3

H²=20²+14²
H²=400+196
H²=596
H=24.41(approx)
H=24.4(exact)

#4

B²=6²-3²
B²=36-9
B²=27
B=5.196(exact)
B=5.2(approx)

3 0

1 year ago

Please solve quickly

Lilit [14]

6/9 .. z. Z. Z. Z. Z. Z.

3 0

2 years ago