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Nesterboy [21]
3 years ago
9

Please help me with these questions

Mathematics
1 answer:
kaheart [24]3 years ago
8 0

Answers:

1st pic:

a) 2

b) 2.5 or 2

2nd pic:

a) 4

b) 5

c) 9

3rd pic:

a) 5

b) 3

Note: I may not be correct on all of them. Some of them were confusing a little bit, so I just wanted to let you know. Please don't come after me in the comments if I am wrong with one of the answers. In reality, you should be doing this. Either way, I hope this helps! Please, please, PLEASE mark brainliest. I would really appreciate it. :)

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Square root of 0.7056
aniked [119]

Answer:

.84

Step-by-step explanation:

√ .7056 put in calculator

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3 years ago
220-109.4 how to solve ?
anastassius [24]

Answer: 110.6

Step-by-step explanation: 220 - 109.4 = 110.6

7 0
3 years ago
Read 2 more answers
Let f(x)=2x+3 and g(x)=x×x-1<br>find a. f(g(x)) b.g(f(x))<br><br>​
Talja [164]

Answer:

See explanations below

Step-by-step explanation:

Given the functions f(x)=2x+3 and g(x)=x^2-1

a. Find f(g(x))

f(g(x)) = f(x^2-1)

f(g(x)) = 2(x^2-1)+3

f(g(x))= 2x^2-2+3

f(g(x)) = 2x^2+1

Hence the composite function f(g(x)) is 2x^2+1

b) g(f(x)) = g(2x+3)

g(f(x) = (2x+3)^2-1

g(f(x)) = 4x^2+12x+9-1

g(f(x)) = 4x^2+12x+8

5 0
2 years ago
Product of 21.33 and 2.04 and the sum of 1.115, 3.18, 22.0613and 12.2384
larisa [96]

Step-by-step explanation:

<u>A</u><u>.</u><u> </u><u>Line up the decimals, then multiply as you would with whole numbers:</u>

21.33 \\\times 2.04 \\  = 43.5132

<u>B. Line up the decimals, then add as you would with whole numbers:</u>

Note: These decimals aren't lined up exactly.

1.115 \\  + 3.18 \\  = 4.295

22.061 \\  + 4.295 \\  = 26.356

26.356 \\  + 12.2384 \\  = 38.5944

7 0
2 years ago
In right ABC, AN is the altitude to the hypotenuse. FindBN, AN, and AC,if AB =2 5 in, and NC= 1 in.
Rama09 [41]

From the statement of the problem, we have:

• a right triangle △ABC,

,

• the altitude to the hypotenuse is denoted AN,

,

• AB = 2√5 in,

,

• NC = 1 in.

Using the data above, we draw the following diagram:

We must compute BN, AN and AC.

To solve this problem, we will use Pitagoras Theorem, which states that:

h^2=a^2+b^2\text{.}

Where h is the hypotenuse, a and b the sides of a right triangle.

(I) From the picture, we see that we have two sub right triangles:

1) △ANC with sides:

• h = AC,

,

• a = ,NC = 1,,

,

• b = NA.

2) △ANB with sides:

• h = ,AB = 2√5,,

,

• a = BN,

,

• b = NA,

Replacing the data of the triangles in Pitagoras, Theorem, we get the following equations:

\begin{cases}AC^2=1^2+NA^2, \\ (2\sqrt[]{5})^2=BN^2+NA^2\text{.}\end{cases}\Rightarrow\begin{cases}NA^2=AC^2-1, \\ NA^2=20-BN^2\text{.}\end{cases}

Equalling the last two equations, we have:

\begin{gathered} AC^2-1=20-BN^2.^{} \\ AC^2=21-BN^2\text{.} \end{gathered}

(II) To find the values of AC and BN we need another equation. We find that equation applying the Pigatoras Theorem to the sides of the bigger right triangle:

3) △ABC has sides:

• h = BC = ,BN + 1,,

,

• a = AC,

,

• b = ,AB = 2√5,,

Replacing these data in Pitagoras Theorem, we have:

\begin{gathered} \mleft(BN+1\mright)^2=(2\sqrt[]{5})^2+AC^2 \\ (BN+1)^2=20+AC^2, \\ AC^2=(BN+1)^2-20. \end{gathered}

Equalling the last equation to the one from (I), we have:

\begin{gathered} 21-BN^2=(BN+1)^2-20, \\ 21-BN^2=BN^2+2BN+1-20 \\ 2BN^2+2BN-40=0, \\ BN^2+BN-20=0. \end{gathered}

(III) Solving for BN the last quadratic equation, we get two values:

\begin{gathered} BN=4, \\ BN=-5. \end{gathered}

Because BN is a length, we must discard the negative value. So we have:

BN=4.

Replacing this value in the equation for AC, we get:

\begin{gathered} AC^2=21-4^2, \\ AC^2=5, \\ AC=\sqrt[]{5}. \end{gathered}

Finally, replacing the value of AC in the equation of NA, we get:

\begin{gathered} NA^2=(\sqrt[]{5})^2-1, \\ NA^2=5-1, \\ NA=\sqrt[]{4}, \\ AN=NA=2. \end{gathered}

Answers

The lengths of the sides are:

• BN = 4 in,

,

• AN = 2 in,

,

• AC = √5 in.

7 0
1 year ago
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