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leva [86]
3 years ago
6

Three brothers Don,Jay ,and Bret—are each one year apart in age.The total of their ages is 27. What are the boys’ ages from youn

gest (Don)to oldest (Brett)?
Mathematics
1 answer:
Romashka-Z-Leto [24]3 years ago
5 0
Don is eight,jay is nine,and,and Bret is 10.

1
10
9
8
➖
27
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If A = (-2, -4) and B = (-8, 4), what is the length of AB?
Oksi-84 [34.3K]

Answer:

10

Step-by-step explanation:

Using the distance formula,

\sqrt{ {( - 2 - ( - 8))}^{2}  +  { (- 4 - 4)}^{2} }  = 10

5 0
2 years ago
What is an irrational number number between 4.5 and 4.6
Marat540 [252]
To 4.5, add the difference between the two numbers (0.1) multiplied by some irrational number greater than 1
3 0
3 years ago
Slove all of it please 100 points
Katyanochek1 [597]

Answer:

Step-by-step explanation:

5x+y-7=0 isolate y

y=-5x+7 (m=-5,b=7)

9x+7y-14=0

7y=14-9x ( divide both sides by 7)

y=-9x/7 +14/7

y=-9/7 + 2  (m=-9/7, b=2)

x+y=1000

4x+3y=3600 ( multiply the first equation by 3 to eliminate y)

3x+3y=3000

4x+3y=3600 ( subtract the two equations)

3x+3y-(4x+3y)=3000-3600

3x+3y-4x-3y=-600

-x=-600

x=600 ( substitute x in the equation)

x+y=1000

600+y=1000

y=1000-600=400

check : x+y=1000 , (600+400)=1000

check : 4x+3y=3600 = (4(600)+3(400))=2400+1200=3600

how many tickets are sold :

adults x=600/4=150

kids (y)=400/3=133 tickets+1 dollars left

unless( x represent kids then x=200, y represent adult = 100)

4 0
3 years ago
Read 2 more answers
Jack took a test and got 13 out of 17 points. He took a make-up test and got 12 out of 15 points.
andrezito [222]
The answer is choice A, a 3.5% increase.
12/15 = 80%
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6 0
3 years ago
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Please Help Quickly!!!<br><br> Find the limit if f(x) = x^3
jeka94

Answer:

Option b. 12

Step-by-step explanation:

This exercise asks us to find the derivative of a function using the definition of a derivative.

Our function is f(x) = x^{3}. Therefore:

f(2+h) = (2+h)^{3}

f(2) = (2)^{3} = 8

Then:

\lim_{h \to \0} \frac{f(2+h)-f(2)}{h}=\lim_{h \to \0} \frac{(2+h)^{3}-8}{h}

Expanding:

\lim_{h \to \0} \frac{(2+h)^{3}-8}{h} =\lim_{h \to \0} \frac{8+ h^{3} +6h(2+h) -8}{h} =\lim_{h \to \0} \frac{h^{3} +6h(2+h)}{h}

\lim_{h \to \0} \frac{h^{3}+ 6h(2+h)}{h} =\lim_{h \to \0} h^{2} + 6(2+h)

Now, if x=0:

\lim_{h \to \0} \frac{f(2+h)-f(2)}{h} = (0)^{2} +6(2+0) = 12

4 0
3 years ago
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