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Svet_ta [14]
3 years ago
15

Help me please and explain if possible?

Mathematics
1 answer:
creativ13 [48]3 years ago
8 0
Let x = months

so we know that price of software package is $20
Now we need to find x the price of one month

we know that Angie and Kenny spent 115 total

Angie: 20 + 3x ( one software pack + 3(price of one month)
Kenny: 20+ 2x ( one software pack + 2( price of one month)

So Angie + Kenny = $ 115

20 + 3y + 20 + 2y = 115

combine like terms and solve for x. That will give you the cost of one month

40 + 5x = 115
-40 -40

5x= 75
5x/5 = 75/5

x=15
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3 divided by 11/12 <br> I need the answer ASAP
s344n2d4d5 [400]

36/11

Calculator solves everything

8 0
3 years ago
Adele opens an account with $100 and deposits $45 a month. Kent opens an account with $70 and also deposits $45 a month. Will th
Harman [31]

answer:

Adele and Kent will not have the same amount in their account at the same time. because they both are going at the same rate depositing 45$a month Adele is ahead by 30$ because she deposited 100 instead of 70 in the beginning. They will be at the same point if Kent deposits more or Adele deposit less.

3 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
How many intersections are there of the graphs of the equations below? x + 5y = 6 3x + 30y = 36 none one two infinitely many How
Aloiza [94]

You can solve this system of the equations with graphical method or algebra method.

I solve this system with Gaus algorithm

first we divide second equation with number 3 and get x+10y=12

Than subtract first equation from second and get 5y=6 => y=6/5

now we replace y in the first equation and get x+5*(6/5)=6 =>x+6=6

=> x=0           (x,y)=(0,6/5)

Conclusion is that we have one intersection point between this two graphs.

Good luck!!!


8 0
4 years ago
Read 2 more answers
I need help with this
Aleksandr [31]

Answer:

3. m∠1 = 106° ~ this is because ∠1 and ∠2 together make a straight line and are therefore supplementary, meaning added together, they equal 180° (so I did 180° - 74° = 106°)

4. m∠3 = 74° ~ again, it is supplementary to ∠1. It is also equal to ∠2

5. m∠8 = 114° ~ angles opposite of each other (like 1 and 4) are equal (as we know from question 4). From there, we can use the corresponding angle theorem, so we know 4 and 8 are congruent. (also you can just know 1 and 8 are congruent by using the opposite exterior angles theorem)

6. m∠6 = 124° ~ using same-side interior angle theorem, they are supplementary angles (or the corresponding angles theorem mentioned above, make 4 congruent to 8, and 8 is supplementary to 6)

7. m∠7 = 96° ~ using same side exterior angle theorem, these angles are supplementary

8. m∠2 = 64° ~ again, same side exterior angle theorem

4 0
2 years ago
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