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3 years ago

Is each statement true for OA? Yes No KA ≈ MA L JKA ≈ L MLA L JKA ≈ L MLA

Mathematics

1 answer:

OLEGan [10]3 years ago

7 0

Yes they are true .

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A student runs 100 meters in 11 seconds. What is the speed of the student? speed = distance over time; 1 mile = 1609 meters

elena-14-01-66 [18.8K]

S=D/T D=100m T=11 100/11=9.1m/s^2

7 0

3 years ago

What is the slope of the graph of y = x^2 + 12 -7 at the point x = 4

Doss [256]

Answer:

Step-by-step explanation:

Step 1: find the derivative of y = x^2 + 12x - 7; it is 2x + 12.

This function represents the slope of the tangent line to the given curve at x.

Step 2: Substitute 4 for x in f '(x) = 2x + 12: f '(4) = 2(4) + 12 = 20

The slope of the tangent line to the graph of y = x^2 + 12x - 7 at x = 4 is 20.

6 0

3 years ago

Solve for x: 3-(2x-5)<-4(x+2)

andriy [413]

3-(2x-5)<-4(x+2)

8-2x<-8-4x

16<-2x

-8 > x. or equivalently x<-8

3 0

4 years ago

Graph a parabola whose x-intercepts are at x = -3 and x = 5 and whose minimum value is y = -4.

shtirl [24]

Answer:

(See explanation and attachment below for further details)

Step-by-step explanation:

The parabola must satisfy the following conditions:

$y + 4 = C\cdot (x - k)^{2}$

$y + 4 = C\cdot (x^{2}-2\cdot k \cdot x + k^{2})$

The expression for the x-intercepts are, respectively:

x = -3

$4 = C\cdot [(-3)^{2} - 2\cdot k\cdot (-3)+k^{2}]$

$\frac{4}{C} = 9 + 6\cdot k + k^{2}$

x = 5

$4 = C\cdot [5^{2}-2\cdot k \cdot (5)+k^{2}]$

$\frac{4}{C} = 25 - 10\cdot k + k^{2}$

By equalizing both expressions:

$9 + 6\cdot k + k^{2} = 25 - 10\cdot k + k^{2}$

$16\cdot k = 16$

$k = 1$

And,

$C = \frac{4}{25-10\cdot (1)+1^{2}}$

$C = \frac{1}{4}$

The equation of parabola is:

$y = \frac{1}{4}\cdot (x-1)^{2} - 4$

Whose graph is included as attachment.

5 0

3 years ago

What’s the product of -2.4(3)

natta225 [31]

-7.2 because you have to multiply-2.4 by 3

4 0

3 years ago

Read 2 more answers