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SVEN [57.7K]
2 years ago
14

Simplify (7-6t)+(8-3i)

Mathematics
2 answers:
Blizzard [7]2 years ago
8 0
Use the power rule
a
m
a
n
=
a
m
+
n
a
m
a
n
=
a
m
+
n
to combine exponents.
−
56
+
21
i
+
48
i
−
18
i
1
+
1
-
56
+
21
i
+
48
i
-
18
i
1
+
1
Add
1
1
and
1
1
to get
2
2
.
−
56
+
21
i
+
48
i
−
18
i
2
-
56
+
21
i
+
48
i
-
18
i
2
Rewrite
i
2
i
2
as
−
1
-
1
.
−
56
+
21
i
+
48
i
−
18
⋅
−
1
-
56
+
21
i
+
48
i
-
18
⋅
-
1
Multiply
−
18
-
18
by
−
1
-
1
to get
18
18
.
−
56
+
21
i
+
48
i
+
18
-
56
+
21
i
+
48
i
+
18
Add
−
56
-
56
and
18
18
to get
−
38
-
38
.
−
38
+
21
i
+
48
i
-
38
+
21
i
+
48
i
Add
21
i
21
i
and
48
i
48
i
to get
69
i
69
i
.
−
tatiyna2 years ago
4 0
<span>Solution) i = {2.666666667, 1.166666667}


</span>
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Express {x+1}{2x-3} as a trinomial.
Marizza181 [45]

Answer:

2x^2 + 5x + 3

Step-by-step explanation:

First, expand the original equation:

(x + 1)(2x + 3)

2x^2 + 3x + 2x + 3

Now simplify:

2x^2 + 5x + 3

And that's your answer!

6 0
3 years ago
A plane intersects a triangular prism perpendicular to its bases.What 2-D cross section is formed
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B. triangle is formed.
3 0
3 years ago
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According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.
UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

<em>Let </em>\hat p<em> = sample proportion of Americans who can order a meal in a foreign language</em>

The z-score probability distribution for sample proportion is given by;

          Z = \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( \hat p > 0.50)

  P( \hat p > 0.06) = P( \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } > \frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } } ) = P(Z > 0.85) = 1 - P(Z \leq 0.85)

                                                               = 1 - 0.80234 = <u>0.19766</u>

<em>Now, in the z table the P(Z  </em>\leq <em>x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.</em>

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0
3 years ago
The school that Imani goes to is selling tickets to the annual dance competition. On the first day of the ticket sales the schoo
Tatiana [17]

Answer:

The adult and the child ticket are both 8 dollars

Step-by-step explanation:

x =adult ticket price

y = child ticket price

I will assume you forget to put that they sold 2 child tickets on the second day

7x+5y=96 and 3x+2y= 40

I will use elimination.  Multiply the first equation by 2 and the second equation by -5 to eliminate y

2(7x+5y)=96*2

14x + 10y = 192

The second equation

-5(3x+2y)= 40*-5

-15x -10y = -200

Add the equations together

14x + 10y = 192

-15x -10y = -200

------------------------

-x = -8

Multiply by -1

x = 8

Now we need to find y

3x+2y= 40

3(8) +2y = 40

24+2y = 40

Subtract 24 from each side

24-24 +2y = 40-24

2y = 16

Divide by 2

2y/2 =16/2

y =8

The adult and the child ticket are both 8 dollars


8 0
3 years ago
Simplify fully <br>2/(5x)+2x/3<br>​
lara31 [8.8K]

Answer:

16x / 15

Step-by-step explanation:

2/5x + 2x/3

6x - 10x /15

16x / 15

7 0
3 years ago
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