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Sergio039 [100]

3 years ago

5

What is -5t+7=11t+25 solved

1 answer:

skad [1K]3 years ago

6 0

The answer is t=2!!!!!!

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Yesterday, Michelle's Snack Shack went through 3 3/5 bottles of ketchup. If they used 1 3/10 times as much mustard as ketchup, h

Dima020 [189]

Okay. To find the answer to this question, let's multiply both fractions together. 3 3/5 as an improper fraction is 18/5 and 1 3/10 as an improper fraction is 13/10. Multiply straight across to get the values. 18/5 * 13/10 is 234/50 or 4 17/25 in mixed number form. Michelle's Snack Shack went through 4 17/25 bottles of mustard.

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2 years ago

Graph the line using a point and a slope. Write the equation of each line. A line that passes through the point (0, –3) and para

kiruha [24]

Y + 3 = 1.2(x - 0)

y + 3 = 1.2x - 0

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2 years ago

Drag the tiles to the correct boxes to complete the pairs not all tiles will be used match each quadratic graph to its respectiv

ch4aika [34]

Answer:

Part 1) The function of the First graph is $f(x)=(x-3)(x+1)$

Part 2) The function of the Second graph is $f(x)=-2(x-1)(x+3)$

Part 3) The function of the Third graph is $f(x)=0.5(x-6)(x+2)$

See the attached figure

Step-by-step explanation:

we know that

The quadratic equation in factored form is equal to

$f(x)=a(x-c)(x-d)$

where

a is the leading coefficient

c and d are the roots or zeros of the function

Part 1) First graph

we know that

The solutions or zeros of the first graph are

x=-1 and x=3

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-3)(x+1)$

Find the value of the coefficient a

The vertex is equal to the point (1,-4)

substitute and solve for a

$-4=a(1-3)(1+1)$

$-4=a(-2)(2)$

$a=1$

therefore

The function is equal to

$f(x)=(x-3)(x+1)$

Part 2) Second graph

we know that

The solutions or zeros of the first graph are

x=-3 and x=1

The parabola open down, so the leading coefficient a is negative

The function is equal to

$f(x)=a(x-1)(x+3)$

Find the value of the coefficient a

The vertex is equal to the point (-1,8)

substitute and solve for a

$8=a(-1-1)(-1+3)$

$8=a(-2)(2)$

$a=-2$

therefore

The function is equal to

$f(x)=-2(x-1)(x+3)$

Part 3) Third graph

we know that

The solutions or zeros of the first graph are

x=-2 and x=6

The parabola open up, so the leading coefficient a is positive

The function is equal to

$f(x)=a(x-6)(x+2)$

Find the value of the coefficient a

The vertex is equal to the point (2,-8)

substitute and solve for a

$-8=a(2-6)(2+2)$

$-8=a(-4)(4)$

$a=0.5$

therefore

The function is equal to

$f(x)=0.5(x-6)(x+2)$

3 0

3 years ago

1 2 3 4 5 6 7 8 9 10

kirza4 [7]

Yes so i do need to doing the times and multipics then get a fraction then u will get tot eh answer

6 0

3 years ago

Help me with my math

aleksley [76]

No nonononononononono

3 0

3 years ago