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Ierofanga [76]
3 years ago
12

Expand.If necessary, combine like terms.(2+7x)(2-7x)stuck? Watch a vide​

Mathematics
2 answers:
m_a_m_a [10]3 years ago
8 0

Answer:4-49x^2

Step-by-step explanation:

2^2-(7x)^2= given ans

scoray [572]3 years ago
3 0

Answer:

\boxed{\bold{-49x^2+4}}

Explanation:

Apply Difference of Two Squares Formula: \bold{\left(a+b\right)\left(a-b\right)=a^2-b^2}

\bold{a=2,\:b=7x}

= \bold{2^2-\left(7x\right)^2}

Simplify \bold{2^2-\left(7x\right)^2: \ -49x^2+4}

\boxed{\bold{-49x^2+4}}

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Write an equation of a line that is perpendicular to the line 2x-3y+15=0
loris [4]

Answer:

y =- \frac{3}{2} x+1

Step-by-step explanation:

The equation of the given straight line is 2x - 3y + 15 = 0.

Now, we can express the equation as 3y = 2x + 15

⇒ y = \frac{2}{3} x+ 5 ..........(1)

The equation is in slope-intercept form and the slope is =\frac{2}{3}

Therefore, the equation of a straight line which is perpendicular to (1) is  

y = -\frac{3}{2} x+c {Where c is any constant}

{Since, the product of slopes of two perpendicular straight lines is -1}

Now, putting c = 1, we get the equation as y =- \frac{3}{2} x+1. (Answer)

4 0
3 years ago
Find the equation of the line that goes through (2,4) and has a slope of -2
AleksAgata [21]

Answer:

y=-2x+8

Step-by-step explanation:

y-y1=m(x-x1)

y-4=-2(x-2)

y-4=-2x+4

y=-2x+4+4

y=-2x+8

3 0
3 years ago
Find an equation of the line that passes through the points (1,2) and (2,3).
klemol [59]

Answer:

y=x+1

Step-by-step explanation:

m=(y2-y1)/(x2-x1)

m=(3-2)/(2-1)

m=1/1

m=1

y-y1=m(x-x1)

y-2=1(x-1)

y-2=x-1

y=x-1+2

y=x+1

6 0
3 years ago
Read 2 more answers
1) If the parent function is f(x)=\xl, which transformation would shift the
gayaneshka [121]

Answer:

The last one.

Step-by-step explanation:

4 0
3 years ago
A cup of coffee with temperature 155degreesF is placed in a freezer with temperature 0degreesF. After 5 ​minutes, the temperatur
Leviafan [203]

Answer:

45.50° F

Step-by-step explanation:

As per Newton's law,

T(t) = T_{s}+(T_{0} + T_{s})e^{-kt}

When T(t) is the final temperature

T_{s} = Temperature of surrounding

T_{0} = Initial temperature

t =duration of cooling

k = constant

103=0+(155-0)e^{-k\times 5}

103=155e^{-5k}

Now take natural log on both the sides

ln(103)=ln(155e^{-5k})

ln(103)=ln(155)+ln(e^{-5k})

ln(103)=ln(155)=-5k

4.6347 - 5.0434 = -5k

k=\frac{0.40872}{5}

k = 0.0817

T(t)=0+(155-0)e^{(0.0817\times 15)}

   = 155e^{-1.226}

   = 155 (02935)

  = 45.49 ≈ 45.50° F

8 0
3 years ago
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