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4 years ago

Help with number 18 please! :)

Mathematics

2 answers:

Korolek [52]4 years ago

7 0

The answer is D all you have to do is add the numbers up the look at the answer choices it’s not that hard

aivan3 [116]4 years ago

3 0

Answer:

The answer is D

Step-by-step explanation:

Take each coordinate set and for every x value add 5 and for every y value subtract 1.

3 +5 = 8, 7-1 = 6 == A'(8,6)

D is the only answer choice with this option so the answer is D.

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A bag contains 5 red marbles, 10 blue marbles, and 14 yellow marbles. If Jen selects a marble from the bag without looking, what

Darina [25.2K]

Answer:

The answer to your question is 0.83 or 83%

Step-by-step explanation:

Data

5 red marbles

10 blue marbles

14 yellow marbles

Process

1.- Calculate the total amount of marbles

Total amount = 5 red + 10 blue + 14 yellow

Total amount = 29 marbles

2.- Calculate the amount of marbles that are not red

No red marbles = 10 + 14

= 24 marbles

3.- Calculate the probability

Probability of not pulling a red marble = $\frac{no red marble}{total amount of marbles}$

Probability of not pulling a red marble = $\frac{24}{29} = 0.83 or 83%$

5 0

3 years ago

♥Please answer soon!♥Find the lateral area of the cone. Use 3.14 pi and round to the nearest whole unit

Tom [10]

I think it’s A, im not sure.

4 0

4 years ago

40 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!! HELPPPPPPPP!!!

Dmitrij [34]

School to mall - 16 miles
mall to library - 30 miles

16²+30²=X²
256+900=X²
X²=1156
X=34

16+30=46
hypotenuse = 34
46-34=12
So Mike travel 12 miles more.

5 0

4 years ago

Read 2 more answers

1. Fifty-eight thousandths

kap26 [50]

fifty-eight thousandths?

4 0

3 years ago

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

1 year ago