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grigory [225]
3 years ago
13

What is the equation for x+1, x+2, x+3

Mathematics
1 answer:
arlik [135]3 years ago
4 0
<span>Hello there.

What is the equation for x+1, x+2, x+3

</span><span>4x</span>+<span>3</span>
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What is the missing statement in this proof?<br><br><br>{question and answer choices in pictures}
Liono4ka [1.6K]

Answer:

B

Step-by-step explanation:

SAS criterion for congruence proves that two triangles are congruent based on their corresponding parts. Since each of the statements before B states parts of these triangles are congruent. B then states that these two triangles are congruent.

4 0
3 years ago
Plot the value(s) on the number line where this function is equal to zero:
dangina [55]

Step-by-step explanation:

if f(x)=0

then

(x+5)=0 and

(x-1)=0

solve for x

x=-5 and x=1

those are the two values where the function equals zero

Hope that helps :)

8 0
3 years ago
To encourage bulk buying a company reduces the list price of $9000 per unit by seven cents times the number of units bought writ
Misha Larkins [42]

Answer:

R(x) = 8999.93x

Step-by-step explanation:

The original price is $9000 per unit. The unit is x, so if you buy x units, you pay 9000x.

The original price function is

R(x) = 9000x

The discount is 7 cents per unit bought, so if you buy x units, the discount is 9x in cents, or 0.09x in dollars. This discount is subtracted from the original price, so the discounted price is

R(x) = 9000x - 0.07x

R(x) = 8999.93x

Answer: R(x) = 8999.93x

7 0
3 years ago
If a right triangle's hypotenuse is 17 units long, and one of its legs is 15 units long, how long is the other leg?
Vesna [10]

Answer:

8 units

Step-by-step explanation:

Hello!

So, there's a formula we can apply to right-angled triangles: Pythagorean's theorem. It states that  c = \sqrt{{a}^2 + b^{2} }, where <em>c</em> is the hypotenuse and <em>a</em> and <em>b </em>are the legs of the triangle.

So, from the problem,  if <em>c </em>= 17 and <em>a </em> = 15, then, we're solving for <em>b</em>. So we'll rewrite the theorem to solve for <em>b</em>.

{c}^2 = {a}^2+{b}^2\\{c}^2-{a}^2={b}^2\\{b} = \sqrt{{c}^2-{a}^2}

Okay, so now we have isolated the theorem for <em>b. Let's </em>plug in our values for <em>c </em>and <em>a</em>.

b = \sqrt{{17}^2-{15}^2}\\b = \sqrt{289-225}\\b = \sqrt{64}\\b = 8

So, using the theorem, we found <em>b</em> = 8. To check our work, let's plug in <em>b</em> and <em>a</em> and solve for <em>c.</em>

<em />c = \sqrt{{a}^2+{b}^2}}\\c = \sqrt{{15}^2+{8}^2}\\c = \sqrt{225+64}\\c = \sqrt{289}\\c = 17\\<em />

So, we got our hypotenuse to equal 17 units, which is correct! So, our <em>b</em> is correct too. Awesome

7 0
3 years ago
Helpppp plss<br><br><br><br><br> Another one
AnnZ [28]

Answer: there’s no one question though

Step-by-step explanation:

5 0
3 years ago
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