find the height of the triangle

Answer:

In March she sent 752 texts

In April she sent 617

Step-by-step explanation:

The phone bill is 35 + additional texts, so in both bills the value relative to the texts are:

72.6 - 35 = 37.6 for March

65.85 - 35 = 30.85 for April

Each texts costs 0.05, so if we divide the value relative to the texts by 0.05 we will have the number of texts she sent:

37.6/0.05 = 752 texts

30.85/0.05 = 617 texts

Answer:

Tb bight by icing I jb Yves j by by by navy by iffy i by by by

Answer:

240

Step-by-step explanation:

We are given a right triangle. Based on the leg rule, the following equation shows how the length of a leg in a right triangle relates with the segments connected to the hypotenuse:

Hypotenuse/leg = leg/part

Where,

Hypo = 400

Leg = x

Part = 144

Substitute

400/x = x/144

Cross multiply

400*144 = x*x

57,600 = x²

√57,600 = x

240 = x

x = 240

Answer:

4

Step-by-step explanation:

Let's simplify step-by-step.

−2z2+4z+2z2

Combine Like Terms:

=−2z2+4z+2z2

=(−2z2+2z2)+(4z)

=4z

Answer:

4z

Hope this helps <3 Need thanks?

Comment /hearthelp

[tex]\\ \sf\longmapsto -2z^2+4z+2z^2[/tex]

[tex]\\ \sf\longmapsto -2z^2+2z^2+4z[/tex]

[tex]\\ \sf\longmapsto (-2+2)z^2+4z[/tex]

[tex]\\ \sf\longmapsto 0z^2+4z[/tex]

[tex]\\ \sf\longmapsto 4z[/tex]

Answer:

A

Step-by-step explanation:

pgt states if the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

Answer:

Does the answer help you?

Answer:

Solution given:

x^5=225

we have

x=[tex] \sqrt[5]{225} [/tex]

x=2.9541

Hello!

[tex] \bf {x}^{5} = 225[/tex]

[tex] \bf x = \sqrt[5]{225} [/tex]

[tex] \bf x ≈2.95418[/tex]

Answer: x ≈ 2,95418

Good luck! :)

Answer:

p = -5

every element in the sequence is created by subtracting 3 from the pervious element.

Step-by-step explanation:

I am not sure you put every necessary information here.

but I'd the visible information is truly everything, then it's is trivial.

the difference between 4 and 1 is ... well, -3. meaning we subtract 3 from 4 to get 1.

we suspect this is the rule and keep trying.

1 -3 = -2

hey it works.

and -8 -3 = -11

hey, also correct.

and the difference between -2 and -8 is -6, and when we place another item in between (p), we cut that in half again to -3. so, it is all consistent.

therefore,

p = -2 -3 = -5

the rule is

an = an-1 - 3

Answer:

A " (1,-2)

B " (4,0)

C " (6,-3)

Step-by-step explanation:

Hope it helped.

° ° °

Answer:

6.8 gallions i believe. im not quite sure

Answer:

the answer is option 1

Step-by-step explanation:

the negative angle => a quarter round angle (clockwise) = ¼ x ( -360°) = -90°

and

the positive angle => a full round angle + 270° = (360°)+270°

= 630°

Answer:

correct answer is -90 and 630

Step-by-step explanation:

Answer:

O D. 3x2 + 2x-3

Answer:

A.7x2-5x questions 2of 20

Answer:

i use logical reasoning

Answer:

hope this help you

have a great day

Answer:

The area of the circle is 100 square units.

Step-by-step explanation:

We are given that the circumference of a circle is 20π, and we want to determine its area.

Recall that the circumference of a circle is given by the formula:

[tex]\displaystyle C = 2\pi r[/tex]

Substitute:

[tex]20 \pi = 2 \pi r[/tex]

Solve for the radius:

[tex]\displaystyle r = \frac{20\pi}{2\pi} = 10[/tex]

The area of a circle is given by:

[tex]\displaystyle A = \pi r^2[/tex]

Since the radius is 10 units:

[tex]\displaystyle A = \pi (10)^2[/tex]

Evaluate:

[tex]\displaystyle A = 100\pi\text{ units}^2[/tex]

In conclusion, the area of the circle is 100 square units.

Answer:

A^7

Step-by-step explanation:

the answer is

A. 28, 25, 22

Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}\\\\csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{hypotenuse}{adjacent}\\\\cot(\theta)=\frac{adjacent}{opposite}[/tex]

Please note that the terms (opposite) and (adjacent) are relative to the angle uses in the ratio, however the term (hypotenuse) refers to the side opposite the right angle, this side never changes its name. Use these ratios to evaluate the trigonometric functions. Then simplify to prove the identity.

2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

Simplify,

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}[/tex]

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

Simplify,

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

Rationalize the denominator,

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

[tex]2-\sqrt{3}=\frac{(\sqrt{3}-1)^2}{3-1}[/tex]

[tex]2-\sqrt{3}=\frac{3-2\sqrt{3}+1}{2}[/tex]

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

Therefore, this equation is also true.

Answer:

1≤f(x)≤5

Step-by-step explanation:

-1≤x≤1

-2≤2x≤2 (Multiplied by 2 both side)

-2+3≤2x+3≤2+3 (Adding three both sides)

1≤f(x)≤5

Answer:

apoco la propiedad asociativa en los siguiente ejercicio 25x11x18=

Answer:

k = 14

Step-by-step explanation:

Prime factorize 32

32 = 2 * 2 * 2 * 2 * 2  = 2⁵

[tex]\frac{2^{12}}{2^{\frac{k}{2}}}= 32\\\\\frac{2^{12}}{2^{\frac{k}{2}}}=2^{5}\\\\2^{12-\frac{k}{2}}=2^{5}[/tex]

Both sides base are same.So, compare exponents

[tex]12-\frac{k}{2}=5\\[/tex]

Subtract 12 from both side

[tex]-\frac{k}{2}=5-12\\\\-\frac{k}{2}=-7\\[/tex]

Multiply both sides by (-2),

[tex](-2)*(-\frac{k}{2})=-7*(-2)\\\\k = 14[/tex]

Answer:

BD = 10 √3

Step-by-step explanation:

ABC ∆ ,

BD ÷ 20 = sin 60

BD = 20 sin 60

BD = (20 √3) / 2

= 10 √3

Step-by-step explanationIf you are reading this say

thank u