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HACTEHA [7]
3 years ago
14

Can anyone help me out with this problem and help me understand it a little better please.

Mathematics
1 answer:
Nadya [2.5K]3 years ago
4 0

So if you look carefully at the problem, you'll notice you have the letter 'g' in multiple places.

If we treat everything to the left of the equal sign as the 'left side of the equation' and everything to the right of the equal sign as the 'right side of the equation', you'll notice the letter g shows up 2 times on the left side, and 1 time on the right side.

In order to work out what 'g' is, we need to get them all on one side. This means we have to simplify the equation a bit by combining all the g's on each side of the equation and then rearranging.

So if we have (g + 4) - 3g on the left side, we can combine g and -3g, which would give us -2g (because g - 3g = -2g). So the left side becomes -2g + 4.

Let's look at the equation again:

-2g + 4 = 1 + g

We still have a g on the right side, so we need to get rid of it and move it to the left. We do this by subtracting g from both sides (which would cancel out the g on the right because g minus g is just = 0).

So the left side becomes -2g + 4 - g (because we're subtracting g from both sides). If we combine the -2g and -g, we get -3g because -2g - g = -3g. The right side becomes just 1, because g - g = 0.

So let's look at the equation again:

-3g + 4 = 1

Let's move the 4 (by subtracting 4 on both sides) to the other side to get closer to just working out g:

-3g + 4 -4 becomes -3g on the left side, and the right side becomes 1 -4 = -3.

So our equation is:

-3g = -3

Now we just need to work out g. We can do this by dividing by -3 on both sides, as this would cancel out the -3 multiplying the g on the left.

The left side becomes -3g ÷ -3 = g. The right side becomes -3 ÷ -3 = 1.

Therefore, we're left with g = 1.

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lord [1]

Answer:

(- 2, - 1 )

Step-by-step explanation:

Given the 2 equations

2x - 3y = - 1 → (1)

x + 4y = - 6 → (2)

Rearrange (2) expressing x in terms of y by subtracting 4y from both sides

x = - 6 - 4y → (3)

Substitute x = - 6 - 4y into (1)

2(- 6 - 4y) - 3y = - 1 ← distribute and simplify left side

- 12 - 8y - 3y = - 1

- 12 - 11y = - 1 ( add 12 to both sides )

- 11y = 11 ( divide both sides by - 11 )

y = - 1

Substitute y = - 1 into (3) for corresponding value of x

x = - 6 - 4(- 1) = - 6 + 4 = - 2

Solution is (- 2, - 1 )

3 0
3 years ago
Three consecutive odd integers are such that the sum of the first and second is 31 less than 3 times the third. Find the integer
Otrada [13]

1st:x

2nd x+2

3rd: x+4

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subtract 2x from each side

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Answer: 21,23,25

3 0
3 years ago
Zelma buys p pounds of bananas for $0.40 per pound she pays the clerk with a $20 bill the clerk subtract the total cost of the b
Sloan [31]
You didnt include the (presumably) multiple choice answers. My guess is that the answer would be along the lines of: 20 - (P times .40)
7 0
3 years ago
BRAINLIEST ASAP!!!!
ivanzaharov [21]
The graph of a quadratic equation is a U-shaped curved called parabola. It is because the parabola makes very easy to find the axis of symmetry to plot selected points and finding the roots and vertex.

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4 0
3 years ago
Two streams flow into a reservoir. Let X and Y be two continuous random variables representing the flow of each stream with join
zlopas [31]

Answer:

c = 0.165

Step-by-step explanation:

Given:

f(x, y) = cx y(1 + y) for 0 ≤ x ≤ 3 and 0 ≤ y ≤ 3,

f(x, y) = 0 otherwise.

Required:

The value of c

To find the value of c, we make use of the property of a joint probability distribution function which states that

\int\limits^a_b \int\limits^a_b {f(x,y)} \, dy \, dx  = 1

where a and b represent -infinity to +infinity (in other words, the bound of the distribution)

By substituting cx y(1 + y) for f(x, y)  and replacing a and b with their respective values, we have

\int\limits^3_0 \int\limits^3_0 {cxy(1+y)} \, dy \, dx  = 1

Since c is a constant, we can bring it out of the integral sign; to give us

c\int\limits^3_0 \int\limits^3_0 {xy(1+y)} \, dy \, dx  = 1

Open the bracket

c\int\limits^3_0 \int\limits^3_0 {xy+xy^{2} } \, dy \, dx  = 1

Integrate with respect to y

c\int\limits^3_0 {\frac{xy^{2}}{2}  +\frac{xy^{3}}{3} } \, dx (0,3}) = 1

Substitute 0 and 3 for y

c\int\limits^3_0 {(\frac{x* 3^{2}}{2}  +\frac{x * 3^{3}}{3} ) - (\frac{x* 0^{2}}{2}  +\frac{x * 0^{3}}{3})} \, dx = 1

c\int\limits^3_0 {(\frac{x* 9}{2}  +\frac{x * 27}{3} ) - (0  +0) \, dx = 1

c\int\limits^3_0 {(\frac{9x}{2}  +\frac{27x}{3} )  \, dx = 1

Add fraction

c\int\limits^3_0 {(\frac{27x + 54x}{6})  \, dx = 1

c\int\limits^3_0 {\frac{81x}{6}  \, dx = 1

Rewrite;

c\int\limits^3_0 (81x * \frac{1}{6})  \, dx = 1

The \frac{1}{6} is a constant, so it can be removed from the integral sign to give

c * \frac{1}{6}\int\limits^3_0 (81x )  \, dx = 1

\frac{c}{6}\int\limits^3_0 (81x )  \, dx = 1

Integrate with respect to x

\frac{c}{6} *  \frac{81x^{2}}{2}   (0,3)  = 1

Substitute 0 and 3 for x

\frac{c}{6} *  \frac{81 * 3^{2} - 81 * 0^{2}}{2}    = 1

\frac{c}{6} *  \frac{81 * 9 - 0}{2}    = 1

\frac{c}{6} *  \frac{729}{2}    = 1

\frac{729c}{12}    = 1

Multiply both sides by \frac{12}{729}

c    =  \frac{12}{729}

c    =  0.0165 (Approximately)

8 0
3 years ago
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