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2 years ago

If the discriminant is equal to 4, there are .. real solutions.

Mathematics

1 answer:

Aleonysh [2.5K]2 years ago

3 0

Answer:

Yes

Step-by-step explanation:

Since, discriminant is positive, therefore solutions will be REAL.

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Jose Rodriguez's checking account had a starting balance of $1,234.56. He wrote a check for $115.12 for plumbing supplies and a

julsineya [31]

The correct answer is: 
$991.19.

Explanation: 
His starting balance is 1234.56.

Subtracting the check for plumbing supplies, we have: 
1234.56 - 115.12 = 1119.44.

Subtracting the check for his loan payment, we have: 
1119.44 - 225.00 = 894.44.

Depositing the 96.75 we have: 
894.44 + 96.75 = 991.19.

8 0

3 years ago

Read 2 more answers

A point that is exterior to an angle is found on the acute side of the angle. true of false?

laiz [17]

Answer:

True

Step-by-step explanation:

"If one of the exterior angle is acute angle, then corresponding interior angle is obtuse angle."

Hope This Helped

7 0

2 years ago

(11 +33 - 4² ) ÷ ( 8 - 4 )

Ratling [72]

Answer:

Step-by-step explanation:

(11 + 33 - 16) / (8 - 4)

(44 - 16) / (4)

28/4

4 0

3 years ago

Read 2 more answers

Identify all pairs of congruent corresponding parts. Then write another congruence statement for the triangles.

astra-53 [7]

SOLUTION

$\begin{gathered} \Delta FED\cong\Delta CBA \\ \angle F\cong\angle C \end{gathered}$ $\begin{gathered} \Delta EFD\cong BCA \\ \angle E\cong\angle B \\ \bar{DE}\cong\bar{AB} \end{gathered}$ $\begin{gathered} \bar{EF}=\bar{BC} \\ \angle D=\angle A \\ \bar{DF}=\bar{AC} \end{gathered}$

4 0

1 year ago

The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f

fenix001 [56]

Answer:

Area of the figure = 31.25π cm²

Perimeter of the figure = (12.5π + 10) cm

Step-by-step explanation:

Area of the given figure = Sum of area of the two semicircle + area of a quarter circle

Area of one quarter circle (CDE) = $\frac{1}{4}(\pi r^{2})$

= $\frac{1}{4}\pi (5)^2$

= 6.25π

Area of one semicircle = 2 × Area of one quarter circle

= 2(6.25π)

= 12.5π

Area of the given figure = 2×12.5π + 6.25π

= 25π + 6.25π

= 31.25π cm²

Perimeter of the figure = Perimeter of two semicircles + perimeter of one quarter circle + measure of radius DE + measure of radius AB

Perimeter of one semicircle = πr

= 5π cm

Perimeter of one quarter circle = 2.5π

Perimeter of the complete figure = 10π + 2.5π + 5 + 5

= (12.5π + 10) cm

4 0

3 years ago