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vagabundo [1.1K]
3 years ago
15

What is 2 equivalent fractions to the fraction 4/7

Mathematics
1 answer:
ryzh [129]3 years ago
4 0
8 Over 14 And 12 over 21
You might be interested in
F(x)=6x^2 +10x−1
KATRIN_1 [288]

Answer: It has two distinct real zeros.

Step-by-step explanation:

The formula that is used to calculate the discriminant of a Quadratic function is the one shown below:

D=b^2-4ac

In this case you have the following Quadractic function provided in the exercise:

f(x)=6x^2 +10x-1

Let's make it equal to 0:

0=6x^2 +10x-1

You can identify that:

a=6\\\\b=10

Knowing these values, you can substitute them into the formula and then evaluate:

D=10^2-4(6)(-1)\\\\D=124

Therefore, since:

 D>0

You can determine that the it has two distinct real roots.

3 0
4 years ago
Given: y = 3x - 4. what is the x-intercept? (0, -4) (0, 4/3) (4/3, 0)
ankoles [38]
X intercept is 4/3 because you plug zeros in for y to find the y intercept
4 0
3 years ago
Graph a line that contains the point (-5, -6) and has a slope of 2/3.
OlgaM077 [116]

Answer:

I think this will help

Step-by-step explanation:

Make a point at(-5,-6) then go up two boxes and right three boxes.

4 0
3 years ago
Read 2 more answers
What is the answer for this question <br>3(a-7)=6a
algol13
The answer to this is -7
4 0
3 years ago
Read 2 more answers
NASA launches a rocket at t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.
Aleks [24]

Answer:

A.) 24.08 seconds

B.) 825.42 metres

Step-by-step explanation:

function of time is given as

h ( t ) = − 4.9 t 2 + 118 t + 115 .

Where a = -4.9, b = 118, c = 115

Let's assume that the trajectory of the rocket is a perfect parabola.

The time t the rocket will reach its maximum height will be at the symmetry of the parabola.

t = -b/2a

Substitute b and a into the formula

t = -118/-2(4.9)

t = 118/9.8

t = 12.041 seconds

Since NASA launches the rocket at t = 0 seconds, the time it will splash down into the ocean will be 2t.

2t = 2 × 12.041 = 24.08 seconds

Therefore, the rocket splashes down after 24.08 seconds.

B.) At maximum height, time t = 12.041s

Substitute t for 12.041 in the function

h ( t ) = − 4.9 t 2 + 118 t + 115

h(t) = -4.9(12.041)^2 + 118(12.041) + 115

h(t) = -4.9(144.98) + 118(12.041) + 115

h(t) = -710.402 + 1420.82 + 115

h(t) = 825.42 metres

Therefore, the rocket get to the peak at 825.42 metres

6 0
3 years ago
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