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saveliy_v [14]
4 years ago
12

What is the range of the function y=2x+1 for the domain 2<=x<=5?

Mathematics
1 answer:
egoroff_w [7]4 years ago
5 0
So domain is the number you can use
range is the output your get from inputting the domain given
so from 2≤x≤5
since it is linear, we can be sure that we only need to test the endpoints of the domain to find the endpoints of the range

sub 2 for x
y=2(2)+1
y=4+1
y=5

sub 5 for x
y=2(5)+1
y=10+1
y=11

so range is from 5 to 11
in interval notation: [5,11]
in other notation 5≤y≤11
or
R={y|5≤y≤11}
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52

Step-by-step explanation:

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6 0
3 years ago
Find value of X in the isosceles triangle shown below
Dima020 [189]

Answer:

x=10

Step-by-step explanation:

Using Pythagorean theorem,

(√13^2-12^2)×2

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8 0
3 years ago
What is the area of the paper used to wrap a can of soup if the can is 11 cm tall and has a diameter of 7.5 cm. Show all formula
Dimas [21]

Notice that a can is a cylinder, which is basically two circles and one rectangle. Now, the question wants area of this rectangle.

You know the width of the rectangle (or the paper), which is 11cm. How do you find the length then?

Notice that the length of the rectangle is also the circumference of the circles. From here, use C = πd where C is circumference and d is diameter. The d is given, which is 7.5 cm. Depending on the question, π can be 3.14, 22/7 or the actual pi.

multiply π and 7.5cm and you get the circumference. Multiply this to 11cm and you get area of the rectangle.

6 0
3 years ago
19 thousand + 7 tens
velikii [3]
19 thousands is 19 x 1000 = 19,000 
7 tens is 7 x 10 = 70

19 thousands plus 7 tens is 19,000 + 70 which is 19,070.
8 0
3 years ago
<img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7By%7D%20%287%5Csqrt%5B3%5D%7B8y%5E2%7D-%5Csqrt%5B3%5D%7By%5E5%7D%20-4y%5Csqrt%
noname [10]

Answer:

\huge\boxed{-y^2+2y}

Step-by-step explanation:

\sqrt[3]y\cdot\left(7\sqrt[3]{8y^2}-\sqrt[3]{y^5}-4y\sqrt[3]{27y^2}\right)\\\\=(\sqrt[3]y)(7\sqrt[3]{8y^2})-(\sqrt[3]y)(\sqrt[3]{y^5})-(\sqrt[3]y)(4y\sqrt[3]{27y^2})\\\\=7\sqrt[3]{(y)(8y^2)}}-\sqrt[3]{(y)(y^5)}-4y\sqrt[3]{(y)(27y^2)}\\\\=7\sqrt{8y^3}-\sqrt{y^6}-4\sqrt{27y^3}\\\\=7\sqrt[3]{2^3y^3}-\sqrt{y^{2\cdot3}}-4\sqrt{3^3y^3}\\\\=7\sqrt[3]{(2y)^3}-\sqrt{(y^2)^3}-4\sqrt{(3y)^3}\\\\=7\cdot2y-y^2-4\cdot3y\\\\=14y-y^2-12y\\\\=-y^2+2y

Used:

a(a+b)=ab+ac\\\\\sqrt[3]{a\cdot b}=\sqrt[3]a\cdot\sqrt[3]b\\\\\sqrt[3]{a^3}=a\\\\(a^n)^m=a^{n\cdot m}

7 0
2 years ago
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