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3 years ago

ANSWER QUICKLY PLEASE. WILL BE AWARDED BRAINLIEST ANSWER

Mathematics

1 answer:

Alexxandr [17]3 years ago

8 0

Area is found by multiplying the length by the width.

Let X = width.

The length is x +3 ( 3 cm longer than the width).

Now you have area = x * x+3

Area is given as 754 square cm.

754 = x * x+3

Simplify:

754 = x^2 +3x

Subtract 754 from both sides:

x^2 +3x - 754 = 0

Factor:

(x -26) (x+29)

Solve for x:

x = 26 and -29.

Since the dimension cannot be negative, x = 26.

The width = 26 cm.

The length = 26 +3 = 29 cm.

Multiply each dimension by 100 to get millimeters:

Width = 260 mm

Length = 290 mm

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The diameter of Jim’s circular flower bed is 10 feet. What is the area, in square feet, of Jim’s flower bed?

Setler79 [48]

The area of a circle is PiR^2
Divide 10 by 2, to get the radius.
Square the radius, 5, which gives you 25.
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4 years ago

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A bottle of water is 2/3 full. You drink the given portion of

Citrus2011 [14]

Answer:

1/2

Step-by-step explanation:

Step 1: You would multiply 2/3 by 3/4.

(Hint: if it is ____ of ____, you would most likely multiply)

Step 2: 2×3/3×4

Step 3: 2×3=6

Step 4: 3×4=12

Step 5: 6/12

Step 6: Simplify 6/12 = 3/6 = 1/2

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2 years ago

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Identify the x-intercepts of the function below f(x)=x^2+12x+24

damaskus [11]

<u>ANSWER: </u>

x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

<u>SOLUTION:</u>

Given, $f(x)=x^{2}+12 x+24$ -- eqn 1

x-intercepts of the function are the points where function touches the x-axis, which means they are zeroes of the function.

Now, let us find the zeroes using quadratic formula for f(x) = 0.

$X=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Here, for (1) a = 1, b= 12 and c = 24

$X=\frac{-(12) \pm \sqrt{(12)^{2}-4 \times 1 \times 24}}{2 \times 1}$

$\begin{array}{l}{X=\frac{-12 \pm \sqrt{144-96}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{48}}{2}} \\\\ {X=\frac{-12 \pm \sqrt{16 \times 3}}{2}} \\\\ {X=\frac{-12 \pm 4 \sqrt{3}}{2}} \\ {X=\frac{2(-6+2 \sqrt{3})}{2}, \frac{2(-6-2 \sqrt{3})}{2}} \\\\ {X=(-6+2 \sqrt{3}),(-6-2 \sqrt{3})}\end{array}$

Hence the x-intercepts of $\mathrm{x}^{2}+12 \mathrm{x}+24=0 \text { are }(-6+2 \sqrt{3}),(-6-2 \sqrt{3})$

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AleksandrR [38]

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Leah cut a 7 inch ribbon into pieces that are 1/2 an inch long. How many pieces of ribbon did she cut?

photoshop1234 [79]

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