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nadya68 [22]
3 years ago
13

MATH :((how do the chart??

exFormula1" title=" - x ^{2} - 9 \geqslant 0" alt=" - x ^{2} - 9 \geqslant 0" align="absmiddle" class="latex-formula">
​

Mathematics
1 answer:
siniylev [52]3 years ago
5 0

Answer:

<h2>no solution</h2>

Step-by-step explanation:

-x^2-9\geq0\qquad\text{change the signs}\\\\x^2+9\leq0\\\\\text{the parabola}\ x^2\ \text{is op}\text{en up and shifted 9 units up. Therefore is whole}\\\text{over the x-axis (only positive values)}.\\\\\bold{CONCLUSION}:\\\\\bold{no\ solution}

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Consider the polynomial 8x^3+2x^2-20x-5 factor
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Answer:

(4x + 1)(2x² - 5)

Step-by-step explanation:

Given

8x³ + 2x² - 20x - 5 ( factor the first/second and third/fourth terms )

= 2x²(4x + 1) - 5(4x + 1) ← factor out (4x + 1) from each term

= (4x + 1)(2x² - 5)

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The temperature was t°F. It fell 8°F and is now 32°F. What was the original temperature?
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Which are the partial products of 3,706 × 4?
Mariulka [41]

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Step-by-step explanation:

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3 years ago
PLs help me on this..........
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Answer:

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3 0
2 years ago
Read 2 more answers
Janeel has a 10-inch by 12-inch photograph. She wants to scan the photograph, then reduce the result
Vaselesa [24]

<u>Complete Question:</u>

Janeel has a 10 inch by 12 inch photograph. She wants to scan the photograph, then reduce the results by the same amount in each dimension to post on her Web site. Janeel wants the area of the image to be one eight of the original photograph. Write an equation to represent the area of the reduced image. Find the dimensions of the reduced image.

<u>Correct Answer:</u>

A) (10-x)(12-x)=15

B) Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

<u>Step-by-step explanation:</u>

a. Write an equation to represent the area of the reduced image.

Let the reduced dimensions is by x , So the new dimensions are

length=10-x\\breadth=12-x

According to question , Area of new image is :

⇒ Area = \frac{1}{8}Length(breadth)

⇒ Area = \frac{1}{8}(10)(12)

⇒ Area = 15

So the equation will be :

⇒ (10-x)(12-x)=15

b. Find the dimensions of the reduced image

Let's solve :  (10-x)(12-x)=15

⇒ 120-10x-12x+x^2=15

⇒ 120-22x+x^2=15

⇒ x^2-22x+105=0

By Quadratic formula :

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

⇒ x = \frac{22 \pm8 }{2}

⇒ x = \frac{22 +8 }{2} , x = \frac{22 -8 }{2}

⇒ x = 15 , x =7

x = 15 is rejected ! as 15 > 10 ! Side can't be negative

⇒ x =7

Therefore, Dimensions are : Length = 10-x = 3 inch , Breadth = 12-x = 5 inch

5 0
3 years ago
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