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3 years ago

How do you set this question up?

Mathematics

1 answer:

aliina [53]3 years ago

7 0

XY : 7 + 7 = b
XZ : 14 + 24 = 38 / 5 = c
YZ : 10 - 20 = b

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Two planes start from Los Angeles International Airport and fly in opposite directions. The second plane starts 1/2 hour after t

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3 years ago

Collin states that 8+(16+9)=(8+16)+9 is always true . Which of the following properties would prove Collin is correct? 1# transi

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Calculate the total surface area of a pipe with inner surface area 1508.5 square cm, outer surface area 2514.2 square cm, and in

algol [13]

Answer:

The total surface area of the pipe is 4123.31 cm²

Step-by-step explanation:

Total surface area (TSA) of a pipe is calculated as;

Outer area of the pipe + Inner area of the pipe + cross section of the two ends

TSA = 2πRL + 2πrL + 2π(R² - r²)

TSA = 2π(RL + rL + R² - r²)

where;

R is the outer radius

r is the inner radius

L is length of the pipe

Determine the length of the pipe

Outer surface area of the pipe is calculated as;

outer surface of the pipe = 2πRL

2514.2 = 2πRL

2514.2 = (2π x 5) L

2514.2 = 31.42 L

L = 2514.2 / 31.42

L = 80.02 cm

Finally, determine the total surface area of the pipe

TSA = 2π(RL + rL + R² - r²)

TSA = 2π(5x80.02 + 3x80.02 + 5² - 3²)

TSA = 2π(400.1 + 240.06 + 16)

TSA = 2π(656.16)

TSA = 4123.31 cm²

Therefore, the total surface area of the pipe is 4123.31 cm²

8 0

4 years ago

Find the zero of each function and state the multiplicity of each zero. Please show all steps.

vodka [1.7K]

Answer:

1. y=(x+3)^3. Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1). Zeros: x=2 multiplicity 2; x=1 multiplicity 1.

3. y=(2x+3)(x-1)^2. Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

Step-by-step explanation:

1. y=(x+3)^3

$y=0\\ (x+3)^3=0\\ \sqrt[3]{(x+3)^3}=\sqrt[3]{0}\\ x+3=0\\ x+3-3=0-3\\ x=-3$

Zero: x=-3 multiplicity 3.

2. y=(x-2)^2 (x-1)

$y=0\\ (x-2)^2(x-1)=0\\ \left \{ {{(x-2)^2=0} \atop {x-1=0}} \right\\ \left \{ {{\sqrt{(x-2)^2} =\sqrt{0} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x-2=0} \atop {x=1}} \right\\ \left \{ {{x-2+2=0+2} \atop {x=1}} \right\\ \left \{ {{x=2} \atop {x=1}} \right.$

Zeros: x=2 multiplicity 2; x=1 multiplicity 1

3. y=(2x+3)(x-1)^2

$y=0\\ (2x+3)(x-1)^2=0\\ \left \{ {{2x+3=0} \atop {(x-1)^2=0}} \right\\ \left \{ {{2x+3-3=0-3} \atop {\sqrt{(x-1)^2} =\sqrt{0} }} \right\\ \left \{ {{2x=-3} \atop {x-1=0}} \right\\ \left \{ {{\frac{2x}{2} =\frac{-3}{2} } \atop {x-1+1=0+1}} \right\\ \left \{ {{x=-\frac{3}{2} } \atop {x=1}} \right.$

Zeros: x=-3/2 multiplicity 1; x=1 multiplicity 2.

4 0

3 years ago