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3 years ago

Which group of numbers is listed from least to greatest?

Mathematics

1 answer:

salantis [7]3 years ago

6 0

The third on is the answer because absolute value is the distance from zero. you would use the numbers -4, 2, 3, 5, 7.

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Find the vertex angle of the figure below B = 40 egrees

natita [175]

Step-by-step explanation:

it's an Equilateral triangle

the base angels are equal

A=100°

B=40°

C=40°

7 0

3 years ago

A guy wire makes a 67° angle with the ground. Walking out 32 feet further from the tower, the angle of elevation to the top of t

kirill [66]

Answer:

20.9 feet

Step-by-step explanation:

The figure attached below can be formed through the information given.

Step 1: Find angle y

All angles of a triangle are equal to 180°

67 + 39 + y = 180

y = 74°

Step 2: Use sin formula to find H

a/sin a = b/sin b

32/sin 74 = b/sin 39

b = 20.9 feet.

The height of the tower is 20.9 feet.

4 0

3 years ago

Can someone please help-

Sindrei [870]

Answer:

5 $\frac{17}{18} }$

Step-by-step explanation:

6 0

3 years ago

3z2−6z for z=5 Please help?!!

ludmilkaskok [199]

Answer:

3z+2= x ; z=5

3(5)+2

15+2

Step-by-step explanation:

5 0

3 years ago

Plz with steps .. it's very hard can anyone plz

liubo4ka [24]

Answer:

Step-by-step explanation:

$\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\frac{0}{0} \\\\we\ can \ use\ Hospital's\ Rule\\\\\\f(x)=\sqrt{2x}-\sqrt{3x-a} \qquad f'(x)=\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} \\\\g(x)=\sqrt{x} -\sqrt{a} \qquad g'(x)=\dfrac{1}{2\sqrt{x}} \\\\\\\displaystyle\ \lim_{n \to a} \dfrac{\sqrt{2x}-\sqrt{3x-a} }{\sqrt{x}-\sqrt{a}} =\lim_{n \to a} \dfrac{\dfrac{2}{2*\sqrt{2x}} -\dfrac{3}{2*\sqrt{3x-a}} }{\dfrac{1}{2\sqrt{x}} }\\\\$

$\displaystyle \lim_{n \to a} \dfrac{2\sqrt{x} }{\sqrt{2x}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}} =\lim_{n \to a} \dfrac{2 }{\sqrt{2}} -\dfrac{3*\sqrt{x} }{\sqrt{3x-a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3*\sqrt{a} }{\sqrt{2a}}\\\\\\=\dfrac{2}{\sqrt{2}} -\dfrac{3}{\sqrt{2}}\\\\\\=-\ \dfrac{1}{\sqrt{2}}\\\\$

7 0

3 years ago