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erastovalidia [21]

3 years ago

5

James has 2 marbles. His friend gives him 3 marbles. which two doubles facts can you use to find out how many marbles James has

now

1 answer:

timofeeve [1]3 years ago

3 0

You could add 2+3=5thats so easy

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I NEED HELP WITH 2 QUESTIONSS

ryzh [129]

1. 75

2. Between 87 and 88

6 0

3 years ago

Look at the triangle show on the right. The Pythagorean Theorem states that the sum of the squares of the legs of a right triang

Vladimir [108]

Answer:

$cos^2\theta + sin^2\theta = 1$

Step-by-step explanation:

Given

$(\frac{b}{r})^2 + (\frac{a}{r})^2$

Required

Use the expression to prove a trigonometry identity

The given expression is not complete until it is written as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = (\frac{r}{r})^2$

Going by the Pythagoras theorem, we can assume the following.

a = Opposite
b = Adjacent
r = Hypothenuse

So, we have:

$Sin\theta = \frac{a}{r}$

$Cos\theta = \frac{b}{r}$

Having said that:

The expression can be further simplified as:

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$

Substitute values for sin and cos

$(\frac{b}{r})^2 + (\frac{a}{r})^2 = 1$ becomes

$cos^2\theta + sin^2\theta = 1$

3 0

4 years ago

Fake Question So It Doesnt Get Taken Down, What is 40992981x9977, Real Question, How do I change my pfp?

DedPeter [7]

go to where the bell is and where it says "ask question" and then do you see your little icon thingy? right there, click it. hit edit profile. and then

it shoulld say this

Edit your profile

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hit profile picture and boom it says upload and all that

7 0

3 years ago

Read 2 more answers

Johnny has graded 20% of his 150 students papers how many papers does he still need to finish grading ?

LiRa [457]

The correct answer is 120 papers.

4 0

3 years ago

Read 2 more answers

Name two solutions for each inequality.

ahrayia [7]

So,

#1: $n \geq 3 \frac{11}{16} + 4 \frac{1}{2}$

Convert to like improper fractions.
$n \geq \frac{59}{16} + \frac{72}{16}$

Add.
$n \geq \frac{131}{16}\ or\ 8 \frac{3}{16}$

So, one solution could be $8 \frac{3}{16}$ .

Another solution could by 9. There is also 10, 11, 12, etc., and all numbers in between.

#2: $k \ \textless \ 6 \frac{2}{5} * 15$

Convert into improper fraction form.
$k \ \textless \ \frac{32}{5} * 15$

Multiply.
$\frac{(2^5)(3)(5)}{5}$

Cross-cancel, and we have our final result.
$(2^5)(3) = 96$
k < 96

96 is not a solution.

95 is a solution.

So is 94, 93, 92, etc, and all numbers in between.

6 0

3 years ago