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maxonik [38]
2 years ago
14

Help me with this, please.

Mathematics
1 answer:
Ira Lisetskai [31]2 years ago
3 0

Answer:

  • - 5 ± 11

Step-by-step explanation:

2x² + 5x - 12 = 0

  • a = 2,  b = 5, c = -12
  • x = (-b ± D)/2a
  • D = √b² - 4ac = √5² -4*2*(-12) = √25+96 = √121 = 11
  • x = (-5 ± 11)/4

Correct choice is the second one or B

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In the figure below, XZ =DF and _X = D.
yuradex [85]

Answer:

Option (4)

Step-by-step explanation:

Given : XZ ≅ DF and ∠X ≅ ∠D

To prove : ΔXYZ ≅ ΔDEF

Statements                               Reasons

1). XZ ≅ DF                                 1). Given

2). ∠X ≅ ∠ D                              2). Given

3). XY ≅ DE                                3). Required information

4). ΔXYZ ≅ ΔDEF                      4). By SAS property of congruence

Therefore, Option (4) will be the answer.

4 0
3 years ago
What is the domain of the function graphed below?
JulijaS [17]

Answer:

Domain: 0, 5

Range: 0, 2

Step-by-step explanation:

The Domain is the first coordinate which is x.

The Range is the second coordinate which is y.

Hope This Helps :)

5 0
3 years ago
Find m∠X. Put the answer below.<br> m∠X =
tia_tia [17]
This is a 90-60-30 triangle so the angle is 30 degrees.
3 0
2 years ago
1.Bobby walked 5 miles and then 1/2 the distance. What is his total distance?
gtnhenbr [62]

Answer:

7.5 miles

Step-by-step explanation:

Half of 5 equals 2.5. Bobby already walked 5 miles so you add the 2.5 since he walked another half of that distance. 5 + 2.5 = 7.5

6 0
3 years ago
Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given poi
lara [203]

For each curve, plug in the given point (x,y) and check if the equality holds. For example:

(I) (2, 3) does lie on x^2+xy-y^2=1 since 2^2 + 2*3 - 3^2 = 4 + 6 - 9 = 1.

For part (a), compute the derivative \frac{\mathrm dy}{\mathrm dx}, and evaluate it for the given point (x,y). This is the slope of the tangent line at the point. For example:

(I) The derivative is

x^2+xy-y^2=1\overset{\frac{\mathrm d}{\mathrm dx}}{\implies}2x+x\dfrac{\mathrm dy}{\mathrm dx}+y-2y\dfrac{\mathrm dy}{\mathrm dx}=0\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{2x+y}{2y-x}

so the slope of the tangent at (2, 3) is

\dfrac{\mathrm dy}{\mathrm dx}(2,3)=\dfrac74

and its equation is then

y-3=\dfrac74(x-2)\implies y=\dfrac74x-\dfrac12

For part (b), recall that normal lines are perpendicular to tangent lines, so their slopes are negative reciprocals of the slopes of the tangents, -\frac1{\frac{\mathrm dy}{\mathrm dx}}. For example:

(I) The tangent has slope 7/4, so the normal has slope -4/7. Then the normal line has equation

y-3=-\dfrac47(x-2)\implies y=-\dfrac47x+\dfrac{29}7

3 0
3 years ago
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