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qwelly [4]
3 years ago
15

PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!PLSSS HELPPPP I WILLL GIVE YOU BR

AINLIEST!!!!!!PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!! PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!PLSSS HELPPPP I WILLL GIVE YOU BRAINLIEST!!!!!!
What is the solution???!!!!

Mathematics
2 answers:
joja [24]3 years ago
7 0
Y=2x+4
0=2x+4
-2x=4
X=-2
marissa [1.9K]3 years ago
5 0

8

because 2x is like 2×4=8

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Find two numbers between 100 and 150 that have a GCF of 24.
tino4ka555 [31]

Answer:

120 and 144

Step-by-step explanation:

Keep adding 24 to 96

8 0
3 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
What is 0.5% of 50? 15 pts!
BaLLatris [955]

Answer:

W = .25

Step-by-step explanation:

We know that is means equals and of means multiply

W = .5% * 50

Changing percent to decimal form

W = .005 * 50

W = .25

3 0
4 years ago
Read 2 more answers
Which inequality is equivalent to −10(x−1)<13?
Nataliya [291]

Answer:

-10<3

Step-by-step explanation:

first distribute the -10 to all the numbers within the parenthesis.

So, -10(x-1)<13 becomes -10x+10<13

Next, combine like terms. then we get -10<3

4 0
3 years ago
Rewrite 163/4 in as many different ways as you can
VladimirAG [237]
There's an infinite number of ways we can rewrite 16 3/4, some of them being:

16 3/4
16 6/8
16 12/16
16 24/32
16.75
67/4
335/20
670/40
etc.

I hope it will help you ;)

4 0
4 years ago
Read 2 more answers
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