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Iteru [2.4K]
2 years ago
15

How many solutions would the following system of equations have​

Mathematics
2 answers:
defon2 years ago
6 0

Answer:

2 solutions

Step-by-step explanation:

The solution to a system of equations is the point where two lines intersect.

Given the information above, we can see the lines intersect at two different points, making the answer 2 solutions.

Have a lovely rest of your day/night, and good luck with your assignments! ♡

user100 [1]2 years ago
3 0

Answer: 2.

Step-by-step explanation: Both lines intersect each other 2 times.

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7 0
3 years ago
The length of a rectangular floor is 4 feet longer than its width w. The area of the floor is 525 ft^2. A) Write a quadratic equ
mina [271]

Answer:

x^21x+25x-525-0

x^21x+25x-525-0xx^2 - 3.7_) +5^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25

x^21x+25x-525-0xx^2 - 3.7_) +5^2(5^2.x - 3.5^2.7X X 5^2 x(x^2-1-(3.7))+5^2(x-(3.7))=0 x(x-27)+5^2(x-21)-0 (x-21)((x(x-21_) + 5^2(x-21)_)=0 X-21 X-215^2(x-21)(x+5^2)=0(x-21)(x+25)=0X-21=0x+25= 0x-21+21=21(x+25)-25=-25x=21x+25-25=-25x= 21x= -25ensiah193

4 0
3 years ago
Can you write the equation for a circle if the given point does not lie on the circle? Explain .
Pachacha [2.7K]

Answer:

No

Step-by-step explanation:

The equation of a circle with center (a,b) and radius r is given as:

{(x - a)}^{2}  +  {(y - b)}^{2}  =  {r}^{2}

If a given point (x,y) does not lie on this circle, it will not satisfy its equation.

This means the distance from the point to the center is not equal to the radius.

It is either less or greater than the radius.

Hence you cannot write the equation of the circle.

3 0
3 years ago
Seth is using the figure shown below to prove Pythagorean Theorem using triangle similarity. In the given triangle ABC, angle A
lana66690 [7]

The relationship between the lengths of the sides of a right triangle are

given by Pythagoras theorem.

  • Part A: <u>ΔABC is similar to ΔADC</u>
  • Part B: ΔABC and ΔADC are similar according <u>AA similarity postulate</u>
  • Part C: <u>DA = 6</u>

Reasons:

Part A:

∠A = 90°

Segment AD ⊥ Segment BC

Location of point D = Side BC

Part A: In triangle ΔABC, we have;

∠A = 90°, ∠B = 90° - ∠C

In triangle ΔADC, we have;

∠ADC = 90°, ∠DAC = 90° - ∠C

∴ <u>ΔABC is similar to ΔADC</u> by Angle-Angle, AA, Similarity Postulate

Part B: The triangles are similar according to <u>AA similarity postulate</u>,

because two angles in one triangle are equal to two angles in the other

triangle and therefore, by subtraction property of equality, the third angle

in both triangles are also equal.

Part C: The length of DB = 9

The length of DC = 4

Required: Length of segment DA

In triangle ΔABD, we have;

∠BDA = 90°= ∠ADC

∠DAC ≅ ∠B by Congruent Parts of Congruent Triangles are Congruent

Therefore;

ΔABD ~ ΔADC by AA similarity, which gives;

\displaystyle \frac{\overline{DA}}{\overline{DC}}  = \frac{\overline{BD}}{\overline{DA}}

\overline{DA}^2 = \overline{DC} \times \overline{BD}

Which gives;

\overline{DA}^2 = 4 × 9 = 36

\overline{DA} = √(36) = 6

\overline{DA}<u> = 6</u>

Learn more here:

brainly.com/question/2269451

3 0
2 years ago
A toy train with a resistance of 10 ohms is powered by four 1.5-volt batteries. What is the power of the toy train?
Tamiku [17]

Answer:

12

Step-by-step explanation:

4 0
3 years ago
Read 2 more answers
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