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dolphi86 [110]
3 years ago
10

Can you please help me. It says to find each product

Mathematics
1 answer:
WITCHER [35]3 years ago
5 0

Answer:

-3/5

Step-by-step explanation:

7 * -3/7 * 1/5

Rewritng

7/7 * -3/5 *1

Simplifying fractions

1 * -3/5 *1

-3/5

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<img src="https://tex.z-dn.net/?f=%28x%5E3%2B4x%5E2-29x-56%29%2F%28x%2B7%29" id="TexFormula1" title="(x^3+4x^2-29x-56)/(x+7)" al
exis [7]

Answer:

The answer is x2 - 3x - 8.

Step-by-step explanation:

5 0
3 years ago
X +2y= 5
Irina18 [472]

Answer:

6

Step-by-step explanation:

subtract 3 from each side

3 0
2 years ago
Prove that the roots of x2+(1-k)x+k-3=0 are real for all real values of k​
masha68 [24]

Answer:

Roots are not real

Step-by-step explanation:

To prove : The roots of x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0 are real for all real values of k ?

Solution :

The roots are real when discriminant is greater than equal to zero.

i.e. b^2-4ac\geq 0b

2

−4ac≥0

The quadratic equation x^2 +(1-k)x+k-3=0x

2

+(1−k)x+k−3=0

Here, a=1, b=1-k and c=k-3

Substitute the values,

We find the discriminant,

D=(1-k)^2-4(1)(k-3)D=(1−k)

2

−4(1)(k−3)

D=1+k^2-2k-4k+12D=1+k

2

−2k−4k+12

D=k^2-6k+13D=k

2

−6k+13

D=(k-(3+2i))(k+(3+2i))D=(k−(3+2i))(k+(3+2i))

For roots to be real, D ≥ 0

But the roots are imaginary therefore the roots of the given equation are not real for any value of k.

6 0
3 years ago
Which equation is equivalent to 5= 4/3 (6y +9)?
vekshin1

Answer:

option C

Step-by-step explanation:

5 = \frac{4}{3}(6y + 9)\\\\5 = \frac{4\times6y}{3} +\frac{4 \times 9}{3}\\\\5= 8y + 12

8 0
2 years ago
A basketball has a circumference of 29.5 inches using 3.14 as an approximation for pie what is the basketball volume to the near
Elena L [17]
It is about 434 cubic inches.

4 0
3 years ago
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