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vlabodo [156]
3 years ago
10

Please help it will help so much

Mathematics
2 answers:
yanalaym [24]3 years ago
8 0

The answer is d. I already did the math on the calculator

Viefleur [7K]3 years ago
6 0

Answer:

8:10

Step-by-step explanation:

If you add the amount of minutes it takes for her to get ready, then add them to 7:15 (55 minutes after 7:15) you get 8:10

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Type SSS, SAS, ASA, SAA, or HL to
Phoenix [80]

Answer:

ASA

Step-by-step explanation:

6 0
2 years ago
How much 50% sugar syrp shoud you add to 20 quarts of 20% syrup to make it 40%
butalik [34]

Answer:

reduce 10 qauts of 50%

Step-by-step explanation:

3 0
4 years ago
Read 2 more answers
I NEED THIS NOW!!
Airida [17]

Answer:

a) x = 6°  b) m∠B = 50°

Explanation:

(a) total angle measure of a triangle is 180°

8x + 2 + 70° + 60° = 180°

8x + 132° = 180°

8x = 180° - 132°

8x = 48°

x = 6°

(b)

m∠B = 8x + 2

m∠B = 8(6) + 2°

m∠B = 48° + 2°

m∠B = 50°

8 0
3 years ago
Read 2 more answers
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
Please help with this math problem
inna [77]

Answer:

There are 2 solutions  x=0   x=4

Step-by-step explanation:

2 | 6-3x| +4 = 16

Subtract 4 from each side

2 | 6-3x| +4 -4= 16-4

2 | 6-3x| = 12

Divide by 2

2/2 | 6-3x| = 12/2

| 6-3x|  = 6

Now set what is inside the absolute value equal to plus and minus what is on the right side

6-3x = 6            or 6-3x = -6

Subtract 6 from each side

6-6-3x = 6-6            or 6-6-3x = -6-6

-3x =0                                  -3x = -12

Divide by -3 on each side

-3x/-3 =0/-3                                  -3x/-3 = -12/-3

x =0                                                    x=4

There are 2 solutions  x=0   x=4

6 0
3 years ago
Read 2 more answers
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