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3 years ago

Solve. 4= 45 + 16 Please help I mad to finish this today

Mathematics

1 answer:

attashe74 [19]3 years ago

8 0

Answer:

It's false, in case that's what you were looking for.

Step-by-step explanation:

4=45+16

4≠61

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Rewrite the expressions using radical notation

Ber [7]

Tip:

You put the number of each equation into the form.

Exponent’s denominator goes outside

Exponent’s numerator goes inside with the number

4 0

3 years ago

What are the solutions of the equation (2x + 3)2 + 8(2x + 3) + 11 = 0

Tom [10]

Answer:

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

Step-by-step explanation:

The given equation is

$(2x+3)^2+8(2x+3)+11=0$

Let us treat this as a quadratic equation in $(2x+3)$ .

where $a=1,b=8,c=11$

The solution is given by the quadratic formula;

$(2x+3)=\frac{-b\pm \sqrt{b^2-4ac} }{2a}$

We substitute these values into the formula to obtain;

$(2x+3)=\frac{-8\pm \sqrt{8^2-4(1)(11)} }{2(1)}$

$(2x+3)=\frac{-8\pm \sqrt{64-44} }{2}$

$(2x+3)=\frac{-8\pm \sqrt{20} }{2}$

$(2x+3)=\frac{-8\pm2\sqrt{5} }{2}$

$(2x+3)=-4\pm \sqrt{5}$

$(2x+3)=-4-\sqrt{5}$ or $(2x+3)=-4+ \sqrt{5}$

$2x=-3-4-\sqrt{5}$ or $2x=-3-4+ \sqrt{5}$

$2x=-7-\sqrt{5}$ or $2x=-7+ \sqrt{5}$

$x=\frac{-7-\sqrt{5}}{2}$ or $x=\frac{-7+ \sqrt{5}}{2}$

5 0

4 years ago

Two pools are being drained. To start, the first pool had 3850 liters of water and the second pool had 4370 liters of water. Wat

SashulF [63]

Answer:

(a)

Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes

Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes

(b)

3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes

Step-by-step explanation:

(a) With x being the minutes after which you want to calculate the amount of water in the pool, to calculate this amount of water you must subtract the water drained from the initial amount of water that the pool contains.

Knowing that, for example, the water from the first pool drains at a rate of 33 liters per minute, then after x minutes, the total water drained will be 33 liters per minute * x minutes. Then:

Amount of water in the first pool (in liters) = 3850 liters - 33 liters per minute * x minutes

Reasoning in the same way:

Amount of water in the second pool (in liters) = 4370 liters - 43 liters per minute * x minutes

(b) Now want to know when the two pools would have the same amount of water. If they have the same amount of water then you can express:

Amount of water in the first pool = Amount of water in the second pool

Replacing the expressions found in (a):

3850 liters - 33 liters per minute * x minutes= 4370 liters - 43 liters per minute * x minutes

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3 years ago

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Delvig [45]

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3 years ago

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What values of x are solutions for x^2 + 2x = 8 ?

mixas84 [53]

Answer:

option A is right.

mark me brainliest

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3 years ago