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Andrew [12]
2 years ago
11

-5/5 x 18/25 pls tell me

Mathematics
1 answer:
I am Lyosha [343]2 years ago
8 0

Answer:

-.72

Step-by-step explanation:

(-5/5)*(18/25)

(-1)*(18/25)

(-1)*(.72)

-.72

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PLSSS HELP PANICKING
devlian [24]

Answer:

<h2>John has $15 and Alex has $33</h2>

Step-by-step explanation:

a systems of equations can be made from the information on the problem

x+y=48

x=2y+3

since x= 2y+3 substitute 2y+3 in the firat equation to get:

(2y+3)+y=48  -> 3y+3=48 -> 3y=45  -> y=15

plug in 15 for y in the second equation to solve for x

x+15 =48 --> x=33

8 0
2 years ago
What is the complete factorization of x^2+4x-45?​
erma4kov [3.2K]

Answer:(x-5)(x+9)

Step-by-step explanation:

You want two numbers that can give you -45 in multiplication and two numbers that can add to 4 and that is -5 and 9.

5 0
3 years ago
Read 2 more answers
Can Someone Please Help Me With These 2 Problems I Tried Looking Up Videos On How To Get It &amp; I'm Just Confused
Lynna [10]

I am attaching an image with the two solutions represented by the blue areas highlighted in your original graphs. Let me know if you have questions.

6 0
2 years ago
Given that sin theta = 1/4, 0→theta→π/2, what<br>. what is the exact value of cos 8?​
natali 33 [55]
<h3>Answer: Choice B \frac{\sqrt{15}}{4}</h3>

===========================================================

Work Shown:

Angle theta is between 0 and pi/2, so this angle is in quadrant Q1.

Square both sides of the given equation

\sin \theta = \frac{1}{4}\\\\\sin^2 \theta = \left(\frac{1}{4}\right)^2\\\\\sin^2 \theta = \frac{1}{16}

Then use the pythagorean trig identity to get

\sin^2 \theta + \cos^2 \theta = 1\\\\\cos^2 \theta = 1-\sin^2 \theta\\\\\cos \theta = \sqrt{1-\sin^2 \theta} \ \ \ \text{cosine is positive in Q1}\\\\\cos \theta = \sqrt{1-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16}{16}-\frac{1}{16}}\\\\\cos \theta = \sqrt{\frac{16-1}{16}}\\\\\cos \theta = \sqrt{\frac{15}{16}}\\\\\cos \theta = \frac{\sqrt{15}}{\sqrt{16}}\\\\\cos \theta = \frac{\sqrt{15}}{4}\\\\

3 0
3 years ago
Can a sequence be both arithmetic and geometric?
artcher [175]
An aritmetic sequence is like this
a_n=a_1+d(n-1) where a1=first term and d=common difference

geometric is a_n=a_1(r)^{n-1} where a1=first term and r=common ratio


can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same

therfor we need to solve d(n-1)=(r)^{n-1}
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?

when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal

the answer is no, a sequence cannot be both aritmetic and geometric
4 0
3 years ago
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