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3 years ago

HEEEELPPPPPPP!!!!!!!

Mathematics

1 answer:

gayaneshka [121]3 years ago

7 0

I’m pretty sure it’s just 10, because $1,100 is next to 10

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Solve for y :3x-3y=12

Molodets [167]

Answer:x=4+y

Step-by-step explanation:

4 0

3 years ago

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Sheila turned on two buzzers at the

kozerog [31]

5 times because 60÷12 = 5

8 0

3 years ago

Please tell me the answer and tell me how to work it out

FrozenT [24]

Well I believe the answer is A. instead of B,C, or D,

This is because you would not need to multiply right away. The problem addresses that John buys 5 more which means to add and if you needed to multiply right away it would have said twice as much or greater to for an increasive value.

Solution: 2×45+5=f

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4 years ago

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The length of a rectangle is 8 cm longer than its width. find the dimensions of the rectangle if its area is 108cm

hram777 [196]

Answer:

$4+2\sqrt{31}\text{ by } -4+2\sqrt{31}$

Or about 15.136 centimeters by 7.136 centimeters.

Step-by-step explanation:

Recall that the area of a rectangle is given by:

$\displaystyle A = w\ell$

Where w is the width and l is the length.

We are given that the length is 8 centimeters longer than the width. In other words:

$\ell = w+8$

And we are also given that the total area is 108 square centimeters.

Thus, substitute:

$(108)=w(w+8)$

Solve for w. Distribute:

$w^2+8w=108$

Subtract 108 from both sides:

$w^2+8w-108=0$

Since the equation is not factorable, we can use the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this case, a = 1, b = 8, and c = -108. Substitute and evaluate:

$\displaystyle \begin{aligned} w&= \frac{-(8)\pm\sqrt{(8)^2-4(1)(-108)}}{2(1)} \\ \\ &=\frac{-8\pm\sqrt{496}}{2}\\ \\ &=\frac{-8\pm4\sqrt{31}}{2} \\ \\ &=-4 \pm 2\sqrt{31} \end{aligned}$

So, our two solutions are:

$w=-4+2\sqrt{31} \approx 7.136 \text{ or } w=-4-2\sqrt{31}\approx -15.136$

Since width cannot be negative, we can eliminate the second solution.

And since the length is eight centimeters longer than the width, the length is:

$\ell =(-4+2\sqrt{31})+8=4+2\sqrt{31}\approx 15.136$

So, the dimensions of the rectangle are about 15.136 cm by 7.136 cm.

5 0

3 years ago

Please solve the equation X^2+3x+3=0

Degger [83]

x² + 3x + 3 = 0

I can tell by looking that there's no easy way to factor the left side,
so we'll have to use the Quadratic formula.

Before doing that, let's just take a look at the discriminant:

( B² - 4AC ) = (9) - (4 x 3 x 1) = (9 - 12) = -3 .
The discriminant is negative, so we know that the roots will be complex conjugates.

Again: x² + 3x + 3 = 0

x = (1/2) [ -3 plus or minus the square root of (-3) ]

x = -3/2 + (1/2) j√3
and
x = - 3/2 - (1/2) j√3

4 0

3 years ago