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liraira [26]
3 years ago
9

Solve the simultaneous equations: 3a-2b=14 12a+9b=39

Mathematics
1 answer:
kirill [66]3 years ago
5 0
Use elimination
Multiply first equation by 4
12a - 8b = 56
12a + 9b = 39
Subtract both
-17b = 17
b = -1
Plug in -1 for b
3a - 2(-1) = 14
3a = 12, a = 4
Final answer: a = 4, b = -1
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Alex has a collection of sports cards. There are 10 cards in the collection, and 1/5 of them are hockey cards. Alex decides to d
Alborosie
If 1/5 of the cards are hockey cards, there are 2 hockey cards. if he wants to double that amount, he would multiply it by 2, so 2•2=4. The fraction of hockey cards is now is 2/5.
4 0
3 years ago
A builder wants to build the bridge whose cross section is shown in the diagram. Two companies offer simple bids on building the
leva [86]

The length of the bridge is the distance from the beginning to the end.

<em>The distance b between each beam is 9ft.</em>

Let:

<em />I \to<em> I-Beam</em>

<em />d_I \to<em> distance between I beam and the bridge</em>

<em />b \to<em> distance between each I beam</em>

<em />

Given that:

I = \frac 34 ft\\

d_I = 3 ft

<em />Length = 55ft\ 6in<em> --- length of the  bridge</em>

<em />

From the diagram (see attachment), there are: 6 I-beams.

So, the length of the 6 I-beams is:

L_1 = 6 \times I

L_1 = 6 \times \frac 34

L_1 = \frac {18}4

L_1 = 4.5ft

There are 2 I-beams beside the bridge

So, the distance between the 2 I-beams and the bridge is:

d_1 =2 \times d_I

d_1 =2 \times 3ft

d_1 =6ft

There are 5 spaces between the I-beams

So, the length of the total spaces is:

L_2 = 5 \times b

L_2 = 5b

The total length is:

Length = L_1 + d_1 + L_2

So, we have:

4.5ft + 6ft + 5b = 55ft\ 6in

Collect like terms

5b = 55ft\ 6in - 4.5ft - 6ft

5b = 44.5ft\ 6in

Convert inches to feet

5b = 44.5ft\ + \frac{6}{12}ft

5b = 44.5ft\ + 0.5ft

5b = 45ft

Divide both sides by 5

b = 9ft

<em>Hence, the distance (b) between each beam is 9ft.</em>

Read more about lengths at:

brainly.com/question/22059747

3 0
3 years ago
All natural numbers are real numbers
docker41 [41]

Answer: True

Step-by-step explanation: Natural numbers are the set of counting numbers and they begin at 1 and go on forever. All natural numbers are real numbers. In fact, the majority of numbers are real numbers such as imaginary numbers.

6 0
3 years ago
Find the value of c so that (x+2) is a factor of the polynomial<br><br> p(x)=3x^4-x^2+cx-2
DedPeter [7]
Using polynomial long division, we get
         
         3x^3+6x^2+11x
         _____________
(x+2) | 3x^4-x^2+cx-2
          -(3x^4+6x^3)
          ____________
             6x^3-x^2+cx-2
          - (6x^3+12x^2)
         _____________
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If you're wondering how I did the long division, what I essentially did was get the first value (at the start, it was 3x^4) and divided it by the first value of the divisor (which in x+2 was x) to get 3x^3 in our example. I then subtracted the polynomial by the whole divisor multiplied by, for example, 3x^3 and repeated the process.

For this to be a perfect factor, (x+2)*something must be equal to (22+c)x-2. Focusing on how to cancel out the 2, we have to add 2 to it. To add 2 to it, we have to multiply (x+2) by 1. However, there's a catch, which is that we subtract whatever we multiply (x+2) by, so we have to multiply it by -1 instead. We still need to cross out (22+c)x. Multiplying (x+2) by -1, we get
(-x-2) but by subtracting the whole thing from something means that we have to add -(-x-2)=x+2 to something to get 0. x+2-x-2=0, xo (22+c)x-2 must equal -x-2, meaning that (22+c)=-1 and c=-23
4 0
3 years ago
If the area of a rectangle is 12cm2 what could be its length and width
jasenka [17]
There are many different combinations.

We just need two numbers that when you multiply them you get 12.

We can have:

Length = 1
Width = 12

Length = 12
Width = 1

Length = 2
Width = 6

Length = 6
Width = 2

Length = 3
Width = 4

Length = 4
Width = 3
7 0
3 years ago
Read 2 more answers
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