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Misha Larkins [42]
3 years ago
5

Help Wegnerkolmp2741o i need ur help!

Mathematics
1 answer:
melamori03 [73]3 years ago
4 0

Answer:

a

Step-by-step explanation:

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Aline's y-intercept is -6 and its slope is 0.What is its equation in slope-intercept form?
notka56 [123]

Answer:

y=0x+-6

Step-by-step explanation:

y=mx+b

6 0
3 years ago
Why is it that when comparing a squares area with a rectangles area, with the SAME PERIMETER, that the square has a greater area
julsineya [31]

Answer:

When finding for the area of a square ,the lenght will be squared automatically leading to a greater value which is greater that of the rectangle which is length multipled by Width

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3 years ago
Solve the formula t=v+k/g for g
Brut [27]
t=v+\frac{k}{g}\\\\v+\frac{k}{g}=t\ \ \ \ |subtract\ v\ from\ both\ sides\\\\\frac{k}{g}=t-v\iff\frac{g}{k}=\frac{1}{t-v}\ \  \ \ |multiply\ both\ sides\ by\ k\neq0\\\\\boxed{g=\frac{k}{t-v}}
8 0
3 years ago
The table shows a student's proof of the quotient rule for logarithms.
nikklg [1K]

Answer:

The error is at step (3) .

The correct step (3) will be,

\log_{b}(\frac {b^{x}}{b^{y}})

= \log_{b}(b^{x - y})   [by using the laws of indices]

All other steps are correct.

Step-by-step explanation:

The error is at the step (3) , because the student has tried to prove the quotient rule of logarithms by using the property i.e., 'The quotient rule of logarithm' itself , i.e. ,by  assuming the property does hold before proving it. So, the proof is fallacious.

The correct step (3) will be,

\log_{b}(\frac {b^{x}}{b^{y}})

= \log_{b}(b^{x - y})   [by using the laws of indices]

All other steps are correct.

5 0
3 years ago
Given a rectangle with length of (2x+9)cm and width of (3x+1)cm.Two squares, each with sides x cm is removed from the rectangle.
il63 [147K]

Answer: The length is 13cm and the width is 7cm

Step-by-step explanation:

For a rectangle of length L and width W, the area is:

A = W*L

In this case we have:

L = (2*x + 9) cm

W=(3*x + 1) cm

Then the area of the rectangle is:

A = (2*x + 9)*(3*x + 1) cm^2

A = (6*x^2 + 2*x + 27*x + 9) cm^2

A = (6*x^2 + 29*x + 9) cm^2

now we remove two squares with sides of x cm

The area of each one of these squares is (x cm)*(x cm)  = x^2 cm^2

Then the area of the figure will be:

area = (6*x^2 + 29*x + 9) cm^2 - (2*x^2 ) cm^2

area = (4*x^2 + 29*x + 9) cm^2

Now we know that the area of this shape is 83 cm^2, then we need to solve:

83 cm^2 = (4*x^2 + 29*x + 9) cm^2

0 =  (4*x^2 + 29*x + 9) cm^2 - 83 cm^2

0 = (4*x^2 + 29*x - 74) cm^2

Then we need to solve:

0 = 4*x^2 + 29*x - 74

Here we can use Bhaskara's equation, the solutions of this equation are given by:

x = \frac{-29 \pm \sqrt{29^2 - 4*4*(-74)}  }{2*4} = \frac{-29 \pm 45}{8}

Then the two solutions are:

x = (-29 - 45)/8 = -9.25  (for how the length and width are defined, we can not have x as a negative number, then this solution can be discarded).

The other solution is:

x = (-29 + 45)/8 = 2

x = 2

Then the length and width of the rectangle are:

Length = (2*2 + 9)cm = 13 cm

Width = (3*2 + 1)cm = 7cm

4 0
2 years ago
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