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Sliva [168]
3 years ago
6

PLease help me!! Each question is worth 25 points please dont get it wrong please!!

Mathematics
1 answer:
MAXImum [283]3 years ago
7 0

Answer:

1. A (85)(0.08)=6.90 then (85)(0.18)=15.30 then 85 + 6.90 + 15.30 =107.10

2. 82.68

Solving;

She was charged $65 before tax. She had to pay a 6% tax (65)(0.06)=3.90 then 65+3.90=68.90

After this she added 20% of the amount after tax rather 20% of 68.90

68.90(0.20)=13.78 then 68.90+13.78=82.68

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For the graph of the inequality x - 2y >4, what is a value of x for a point that is on the
Colt1911 [192]

Answer:

Step-by-step explanation:

Explanation:

First, find two points on the line if you change the inequality to an equation.

For  

x

=

0

:  

0

+

2

y

=

4

2

y

=

4

2

y

2

=

4

2

y

=

2

or  

(

0

,

2

)

For  

y

=

0

:  

x

+

0

=

4

x

=

4

or  

(

4

,

0

)

We can plot these two points and draw a line through them to get the border of the inequality:

graph{(x^2+(y-2)^2-0.075)((x-4)^2+y^2-0.075)(x+2y-4)=0}

The line will be solid because the inequality operator has a "or equal to" clause in it. We can now shade the area to the right of the line because the inequality has a "great than" clause in it.

graph{(x+2y-4)>=0}

4 0
2 years ago
Compute: 2 3/4 ÷(1 1/2 − 2/5 ) PLZ HELP
ollegr [7]

Answer:

\frac{5}{2}

I hope I helped you! If my answer is correct, please mark me brainiest. Thanks!

8 0
2 years ago
Find the slope of the line hurry!​
Furkat [3]

Answer:

the slope is 3 :)

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
How would you solve 4n-5<-9n+47 to get the solution?
fredd [130]

Answer:

n<4

Step-by-step explanation:

4n-5<-9n+47

13n<52 |:13

n<4

4 0
3 years ago
The vector wequalsaiplusbj is perpendicular to the line axplusbyequalsc and parallel to the line bxminusayequalsc. It is also tr
Likurg_2 [28]

Answer:

\theta=45^{\circ}

Step-by-step explanation:

We are given that the equation of lines

x+\sqrt 3y=1

(1-\sqrt 3)x+(1+\sqrt 3)y=8

According to question

The vector perpendicular to the lines is given by

i+\sqrt 3j and (1-\sqrt 3)i+(1+\sqrt 3)j

Therefore, the  angle between two vectors is given by

cos\theta=\frac{a_1a_2+b_1b_2}{\sqrt{a^2_1+b^2_1}\sqrt{a^2_2+b^2_2}}

Using the formula

cos\theta=\frac{1(1-\sqrt 3)+\sqrt 3(1+\sqrt 3)}{2\times 2\sqrt 2}

cos\theta=\frac{1-\sqrt 3+\sqrt 3+3}{4\sqrt 2}=\frac{1}{\sqrt 2}

cos\theta=cos 45^{\circ}

\theta=45^{\circ}

Hence, the acute angle between the lines is given by

\theta=45^{\circ}

3 0
3 years ago
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