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salantis [7]
2 years ago
9

Ok last one for today that I cant figure out (6c - 4) - ( c - 3) c = -7 Ty!

Mathematics
1 answer:
horrorfan [7]2 years ago
8 0
9
c
−
4
−
c
2
=
−
7
9
c
-
4
-
c
2
=
-
7
Move
7
7
to the left side of the equation by adding it to both sides.
9
c
−
4
−
c
2
+
7
=
0
9
c
-
4
-
c
2
+
7
=
0

Add
−
4
-
4
and
7
7
.
9
c
−
c
2
+
3
=
0
9
c
-
c
2
+
3
=
0
Factor
−
1
-
1
out of
9
c
−
c
2
+
3
9
c
-
c
2
+
3
.
Tap for more steps...
−
(
c
2
−
9
c
−
3
)
=
0
-
(
c
2
-
9
c
-
3
)
=
0
Multiply each term in
−
(
c
2
−
9
c
−
3
)
=
0
-
(
c
2
-
9
c
-
3
)
=
0
by
−
1
-
1
Tap for more steps...
c
2
−
9
c
−
3
=
0
c
2
-
9
c
-
3
=
0
Use the quadratic formula to find the solutions.
−
b
±
√
b
2
−
4
(
a
c
)
2
a
-
b
±
b
2
-
4
(
a
c
)
2
a
Substitute the values
a
=
1
a
=
1
,
b
=
−
9
b
=
-
9
, and
c
=
−
3
c
=
-
3
into the quadratic formula and solve for
c
c
.
9
±
√
(
−
9
)
2
−
4
⋅
(
1
⋅
−
3
)
2
⋅
1
9
±
(
-
9
)
2
-
4
⋅
(
1
⋅
-
3
)
2
⋅
1
Simplify.
Tap for more steps...
c
=
9
±
√
93
2
c
=
9
±
93
2
The final answer is the combination of both solutions.
c
=
9
+
√
93
2
,
9
−
√
93
2
c
=
9
+
93
2
,
9
-
93
2
The result can be shown in multiple forms.
Exact Form:
c
=
9
+
√
93
2
,
9
−
√
93
2
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Answer:

Identify the slope, m. This can be done by calculating the slope between two known points of the line using the slope formula.

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Samantha and Mia each left Julia’s house at the same time. Mia walked north at 7 kilometers per hour. Samantha ran west at 11 ki
Lena [83]
Mia walked 7km/h so after 1 hour, she is 7 km north of the house of Julia

Samantha walked 11km/h so after 1 hour, she is 11 km west of the house of Julia.

The points where Mia and Samantha are after 1 hour , and the house of Julia form a right triangle with sides 7 and 11 km. The distance between the girls, is the hypotenuse of his triangle.

 by the pythagorean theorem:

MS= \sqrt{ 7^{2} + 11^{2} }= \sqrt{49+121}= \sqrt{170}=  13 (km)


Answer: 13 km

3 0
3 years ago
Read 2 more answers
A computer can be classified as either cutting dash edge or ancient. Suppose that 86​% of computers are classified as ancient. ​
Leno4ka [110]

Answer:

(a) The probability that both computers are ancient is 0.7396

(b) The probability that all seven computers are ancient is 0.3479

(c) The probability that at least one of seven randomly selected computers is cutting dash edge is 0.6520.

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

Step-by-step explanation:

We know that 86​% of computers are classified as ancient. This means, if one computer is chosen at random, there is an 86% chance that it will be classified as ancient.

P(ancient)=0.86

(a) To find the probability that two computers are chosen at random and both are ancient​ you must,

The probability that the first computer is ancient is P(ancient)=0.86 and the probability that the second computer is ancient is P(ancient)=0.86

These events are independent; the selection of one computer does not affect the selection of another computer.

When calculating the probability that multiple independent events will all occur, the probabilities are multiplied, this is known as the rule of product.

Let A be the event "the first computer is ancient" and B the event "the second computer is ancient".

P(A\:and \:B)=P(A)\cdot P(B)=0.86\cdot 0.86=0.86^2= 0.7396

(b) To find the probability that seven computers are chosen at random and all are ancient​ you must,

Following the same logic in part (a) we have

Let A be the event "the first computer is ancient",

B the event "the second computer is ancient",

C the event "the third computer is ancient",

D the event "the fourth computer is ancient",

E the event "the fifth computer is ancient",

F the event "the sixth computer is ancient", and

G the event "the seventh computer is ancient"

P(A\:and \:B\:and \:C\:and \:D\:and \:E\:and \:F\:and \:G)=\\P(A)\cdot P(B)\cdot P(C)\cdot P(D)\cdot P(E)\cdot P(F)\cdot P(G) =(0.86)^7=0.3479

(c) To find the probability that at least one of seven randomly selected computers is cutting dash edge​ you must

Use the concept of complement. The complement of an event is the subset of outcomes in the sample space that are not in the event.

Let C the event "the computer is cutting dash edge".

Let A the event "the seven computers are ancient".

P(C)=1-P(A)=1-0.3479=0.6520

Because the probability is about 65% is it not unusual that at least one of seven randomly selected computers is cutting dash edge, it's more likely than not.

5 0
2 years ago
What is 1 and 1/4% of 200?
arsen [322]
5/4 • 200/1, 1000/4
which is 250

250 is 1 1/4 of 200
4 0
2 years ago
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