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tester [92]
3 years ago
10

Find m<1 and m<2, showing ALL work AND

Mathematics
1 answer:
mestny [16]3 years ago
5 0

Answer:

see explanation

Step-by-step explanation:

∠1 = 58° ( alternate angles )

∠1 and ∠2 form a straight angle and are supplementary, thus

∠2 = 180° - 58° = 122°

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4(x+8)-2(x-7)=46 x=?
Scorpion4ik [409]

Answer:

x = 0/2

x = 0

Step-by-step explanation:

4(x+8)-2(x-7) = 46

Expanding the brackets

4x +32 - 2x +14 = 46

2x + 46 = 46

2x = 46-46

2x = 0

x = 0/2

x = 0

5 0
3 years ago
What is the range of the function f(x) = |x| +2?
kykrilka [37]
Imagine the graph of f(x)=|x|. This can only output positive values (x \geq 0).

Then f(x)=|x|+2 is the same graph shifted upwards by two units meaning it can only output values bigger or equal to than 2. So the range is x \geq 2
5 0
4 years ago
Find the product of the given complex number and its conjugate.<br> 1/5+4i
nasty-shy [4]

Answer:

-15.96

Step-by-step explanation:

A conjugate is a binomial with the sign inside changed. So the conjugate of (1/5 + 4i) is (1/5 - 4i)

Set the original and the conjugate next to each other and F.O.I.L. Multiply the first numbers of each binomial, the 1/5 and the 1/5 to get 1/25. This is the "F."

Multiply the outer members, the 1/5 and the 4i to get - 60i. This is the "O."

Multiply the inner  numbers  ( the + 4i and the 1/5) to get + 60i. This is the "I."

Multiply the positive 4i and the negative 4i to get 16i squared

The positive 60i and the negative 60i cancel each other out.

The i squared changes into - 1. This makes the 16 negative.

Add 1/25 to - 16 to get - 15.96

4 0
3 years ago
Type an equivalent expression to show the relationship of multiplication and addition<br> 7+7+7+7+7
nikitadnepr [17]
7 • 5



Multiplication is basically repeated addition. So to convert repeated addition to multiplication, we take the number being added (7) and multiply it be the number of addens (5). Therefore, 7+7+7+7+7 = 5 • 7
3 0
3 years ago
HELP ASAP
astra-53 [7]

Answer:

Step-by-step explanation:

We are given a circle with a partially shaded region. First, we need to determine the area of the whole circle. To do this, we need the measurement of the radius of the circle:

Use the Pythagorean theorem to solve for the other leg of the right triangle inside the circle:

5^2 = 3^2 + x^2

x = 4

The radius is 4 + 1 cm = 5 cm

So the area of the circle is A = pi*r^2

A = 3.14 * (5)^2  

A = 25pi cm^2

To solve for the area of the shaded region:

Ashaded = Acircle - Atriangles

we need to solve for the area of the triangles:

A = 1/2 *b*h

A = 1/2 *6 * 5

A = 15 cm^2

Atriangles = 2 * 15  

Atriangles = 30 cm^2

Ashaded = 25pi - 30

8 0
3 years ago
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