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AVprozaik [17]
4 years ago
15

How do i solve 4m^2+3m-1=0 using the quadratic formula

Mathematics
1 answer:
elixir [45]4 years ago
8 0
Quadratic formulat is used when a^2+bx+c=0 x= \frac{-b+/- \sqrt{b^{2}-4ac}   }{2a} 4m^2+3m-1=0 a=4 b=3 c=-1 input x= \frac{-3+/- \sqrt{3^{2}-4(4)(-1)}   }{2(4)} x= \frac{-3+/- \sqrt{9+16}   }{8} x= \frac{-3+/- \sqrt{25}   }{8} x= \frac{-3+/-5  }{8} x= \frac{-8}{8} or \frac{2}{8} x=-1 or 1/4
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The length of a rectangle is 4cm longer than the width, and the perimeter is at least 48cm. What are the smallest possible dimen
jenyasd209 [6]
So since the length is 4+w.  Then the perimeter 2(4+w) + 2w >= 48.  
so, 8+2w+2w>=48.
8+4w>=48
4w>=40
w>=10
so, the width must be at least 10.  And the smallest dimensions can be 10 x 14.
3 0
3 years ago
What is the simplified form of expression <br><br> 5(14-2)^2/2<br><br> Show your work
Reptile [31]

Answer:

360

Step-by-step explanation:

Order of operations: PEMDAS

Parenthesis

Exponents

Multiply/Divide

Add/Subtract

If same category, go in order.

(14-2)=12

12^2=144

144*5=720

720/2=360

Hope that helped!

7 0
3 years ago
Two equations are shown in the picture please answer. giving brainiest
Dvinal [7]

Answer:well for the first one I am not sure but I think that since four is over the x there should be 4/x + 3 so tragically I don’t know the answer but I think it should be 12 the x should be 12 because 4x3=12 so yea 12

Step-by-step explanation:and for the other one I don’t know but ima try it should be two because 2x2=4 and so y=1-2x2-4 that’s what I think sorry. I got it wrong and do good on your test for me please

6 0
3 years ago
Answer this math question for 15 points please
Sergio [31]

Answer:

.......

Step-by-step explanation:

5 0
3 years ago
What is an equation of the line perpendicular to y=-x-2 and through (-2, 4)?
Salsk061 [2.6K]

Answer: y = x + 6

Step-by-step explanation:

\mathrm{Find\:the\:line\:}\mathbf{y=mx+b}\mathrm{\:perpendicular\:to\:}y=-x-2\mathrm{\:that\:passes\:through\:}\left(-2,\:4\right)

\mathrm{For\:a\:line\:equation\:for\:the\:form\:of\:}\mathbf{y=mx+b}\mathrm{,\:the\:slope\:is\:}\mathbf{m}

m=-1

\mathrm{The\:perpendicular\:slope\:is\:the\:negative\:reciprocal\:of\:the\:given\:slope}

\left(-1\right)m_p=-1

\frac{\left(-1\right)m_p}{-1}=\frac{-1}{-1}

m_p=1

\mathrm{Compute\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:for\:slope\:m=}1\mathrm{\:and\:passing\:through\:}\left(-2,\:4\right)\mathrm{Plug\:the\:slope\:}1\mathrm{\:into\:}y=mx+b

y=x+b

\mathrm{Plug\:in\:}\left(-2,\:4\right)\mathrm{:\:}\quad \:x=-2,\:y=4

4=\left(-2\right)+b

-2+b=4

\mathrm{Add\:}2\mathrm{\:to\:both\:sides}

-2+b+2=4+2

\text{Simplify}

b=6

\mathrm{Construct\:the\:line\:equation\:}\mathbf{y=mx+b}\mathrm{\:where\:}\mathbf{m}=1\mathrm{\:and\:}\mathbf{b}=6

y=x+6

3 0
3 years ago
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