1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
olga2289 [7]
3 years ago
10

Tangent please help me

Mathematics
1 answer:
atroni [7]3 years ago
6 0
A^2+ b^2= c^2
24^2+b^2=25^2
576+b^2=625
Sub 576
b^2= 49 b=7
You might be interested in
3) The half of (√200 + √128 ) is
Sliva [168]

Answer:

Option 2 i.e. 9√2 is the correct option.

Step-by-step explanation:

Given the expression

(\sqrt{200}\:+\:\sqrt{128})

The half of (\sqrt{200}\:+\:\sqrt{128}) can be determined by dividing the expression in half.

Therefore, we need to solve the expression such as

\frac{\left(\sqrt{200}\:+\:\sqrt{128}\:\right)}{2}

as

\sqrt{200}=\sqrt{100\times 2}=\sqrt{100^2\times \:2}=10\sqrt{2}

\sqrt{128}=\sqrt{64\times 2}=\sqrt{8^2\times \:2}=8\sqrt{2}

so the expression becomes

\frac{\left(\sqrt{200}\:+\:\sqrt{128}\:\right)}{2}=\frac{10\sqrt{2}+8\sqrt{2}}{2}

Add similar elements:  10\sqrt{2}+8\sqrt{2}=18\sqrt{2}

                     =\frac{18\sqrt{2}}{2}

Divide the numbers: 18/2 = 9

                      =9\sqrt{2}

Therefore, we conclude that:

\frac{\left(\sqrt{200}\:+\:\sqrt{128}\:\right)}{2}=9\sqrt{2}

Hence, option 2 i.e. 9√2 is the correct option.

4 0
3 years ago
Multiply.
siniylev [52]
The correct answer is A.
4 0
4 years ago
Solve –19 = x – 4. A. x = –23 B. x = –76 C. x = –15 D. x = 25
Marrrta [24]
If you're Looking for X, It's -15. I hope I helped.
6 0
4 years ago
Read 2 more answers
What is the GCF of 18 and 54?
Alchen [17]

Answer:

gcd(18, 54) = 18

Hope this helps have a great day/night :)

7 0
3 years ago
Read 2 more answers
Recall the scenario about Eric’s weekly wages in the lesson practice section. Eric's boss has been very impressed with his work.
Ira Lisetskai [31]

Answer:

f(x)= 12x   where x is the number of hours worked (only whole numbers)

Domain= {10,11,12,13,.............., 25}

Range = { $ 120, $ 132,  $ 144,$ 156 ,$ 168, $ 180, $ 192, $ 204, $ 216, $ 228,        $ 240, $ 252, $ 264, $ 276, $ 288, $ 300}

Step-by-step explanation:

Let x be the number of hours worked then the function will be

f(x)= 12 (x)       for {x=10,11,12,13,.............., 25}

x cannot taken any fractional value or decimal value.

The domain of the function is the input values

Domain= {10,11,12,13,.............., 25}

Now range of the function is the set of all possible output values

f(10)= 12 *10= $ 120

f(11)=  12 *11= $ 132

f(12)=  12 *12= $ 144

f(13)=  12 *13= $ 156

f(14)=  12 *14= $ 168

f(15)=  12 *15= $ 180

f(16)=  12 *16= $ 192

f(17)=  12 *17= $ 204

f(18)=  12 *18= $ 216

f(19)=  12 *19= $ 228        

f(20)=  12 *20= $ 240

f(21)=  12 *21= $ 252

f(22)=  12 *22= $ 264

f(23)=  12 *23= $ 276

f(24)=  12 *24= $ 288

f( 25) = 12*25= $ 300

Range = { $ 120, $ 132,  $ 144,$ 156 ,$ 168, $ 180, $ 192, $ 204, $ 216, $ 228,        $ 240, $ 252, $ 264, $ 276, $ 288, $ 300}

6 0
3 years ago
Other questions:
  • Show how to make a ten to solve 13-7. Write the number sentence
    10·1 answer
  • I WILL LOVE YOU FOREVER. Need answer asapppp
    10·2 answers
  • Which of the following describes two perpendicular number lines that intersect at 0?
    13·1 answer
  • What is 100 times 15
    13·2 answers
  • What is the value of y?
    6·1 answer
  • What is a two step equation that involves multiplication and subtraction includes a negative coefficient and has a solution of x
    15·1 answer
  • The measures of the angles of a triangle are shown in the figure below. Solve for x. ​
    12·1 answer
  • Tarzan (whose mass is 90.0 kg) plans to swing down, startingfrom rest, from a ledge using a light vine 7.4 m long. He holdsone e
    13·1 answer
  • How do I answer this ? (1 + 3x)º<br> 44°
    10·1 answer
  • What’s does 120$ plus 40% mean
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!