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Elis [28]
2 years ago
12

Answer the following equations:

Mathematics
1 answer:
Andrew [12]2 years ago
7 0

Step-by-step explanation:

3(t+5) = 9

3t+15 = 9

3t = -6

t = -2

2(f-7) = -10

2f - 14 = -10

2f = 4

f = 2

-(c - 9) = 4

-c + 9 = 4

-c = -5

c = 5

-6(2t + 8) = -84

-12t - 48 = -84

-12t = -36

12t = 36

t = 3

-10 (s + 2) = -57

-10s - 20 = -57

-10s = -37

s = 3.7

7(3w + 8)/3 = -9

By cross multiplication,

7(3w + 8) = -27

21w + 56 = -27

21w = -83

w = -3.95

35/5 = (F - 32)/9

7 = (F - 32)/9

By cross multiplication,

63 = F - 32

63 + 32 = F

95 = F

<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em>

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Please Help! Show all the steps!
likoan [24]

You can start by subtracting different equations from each other.

3x + 2y + 3z = 1

subtract

3x + 2y + z = 7

2z = -6

divide by 2

z = -3

add the following two expressions together:

3x + 2y + z = 7

3x + 2y + 3z =1

6x + 4y + 4z = 8

subtract the following two expressions:

6x + 4y + 4z = 8

5x + 5y + 4z = 3

x - y = 5

^multiply the whole equation above by 3

3x - 3y = 15

subtract the following two expressions:

3x - 3y = 15

3x + 2y = 10

-5y = 5

divide each side by -5

y=-1

take the following expression from earlier:

x - y = 5

substitute y value into above equation

x - - 1 = 5

2 negatives make a positive

x + 1 = 5

subtract 1 from each side

x = 4

Therefore x = 4, y = -1, z = -3

I checked these with all 3 equations and they worked :)

(it's quite complicated, comment if you don't understand anything) :)

7 0
3 years ago
What is the coefficient of the x5y5-term in the binomial expansion of (2x – 3y)10? 10C5(2)5(3)5 10C5(2)5(–3)5 –10C5(2)5(–3)5 10C
butalik [34]

ANSWER

10C_5(2)^{5}( - 3)^5

EXPLANATION

The given binomial expansion is:

{(2x - 3y)}^{10}

Compare this to

{(a + b)}^{n}

we have a=2x , b=-3y and n=10

We want to find the coefficient of the term

{x}^{5}  {y}^{5}

This implies that, r=5.

The terms in the expansion can be obtained using

T_{r+1}=nC_ra^{n-r}b^r

We substitute the given values to obtain;

T_{5+1}=10C_5(2x)^{10-5}(  - 3y)^5

T_{6}=10C_5(2x)^{5}(3y)^5

T_{6}=10C_5(2)^{5}( - 3)^5 {x}^{5}  {y}^{5}

Hence the coefficient is;

10C_5(2)^{5}( - 3)^5

5 0
4 years ago
Read 2 more answers
17.25 round up to ones
Irina18 [472]
17.24

Since 7 comes after 5 we need to round to the next number.....18.00

17.24 rounded to ones = 18.00

Hope I helped:P

7 0
3 years ago
Read 2 more answers
Q and R are independent events. P(Q) = 0.4; P(Q AND R) = 0.08. Find P(R).
algol [13]
Because Q and R are independent, you have

\mathbb P(Q\cap R)=\mathbb P(Q)\times\mathbb P(R)
0.08=0.4\times\mathbb P(R)
\implies \mathbb P(R)=0.2
8 0
3 years ago
Which is greater 3/8 or 3/4
hram777 [196]
3\4
because the spaces are bigger draw a model and you will see
4 0
3 years ago
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