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2 years ago

Find the value of b that makes the system of equations have the solution (3 ,5).

Mathematics

1 answer:

dexar [7]2 years ago

4 0

$5=b\cdot3+2\\
3b=3\\
b=1 \Rightarrow A$

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The width of the rectangle is 0.6 meters less than the length. The perimeter of a rectangle is 60.8 meters. Find the dimensions

Anvisha [2.4K]

Answer:

14+=+2L+%2B+2%28L-3%29 Simplify and solve for L

14+=+2L+%2B+2L+-+6 Combine like-terms.

14+=+4L-6 Add 6 to both sides.

20+=+4L Divide both sides by 4.

5+=+L The length is 5 meters.

W+=+L-3

W+=+5-3

W+=+2meters.

The width is 2 meters.

P+=+2L%2B2W

P+=+2%285%29%2B2%282%29

P+=+10%2B4

P+=+14meters.

Step-by-step explanation:

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2 years ago

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2 years ago

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Marta_Voda [28]

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2 years ago

Graph a line with a y-intercept of 3 and containing the point (2, 5).

LenKa [72]

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2 years ago

Sec squared 55 - tan squared 55

sergejj [24]

<u>Answer: </u>

sec squared 55 – tan squared 55 = 1

<u>Explanation:</u>

Given, sec square 55 – tan squared 55

We know that,

$\sec \Theta=\frac{\text {hypotenuse}}{\text {base}}$

And,

$\tan \theta=\frac{\text { perpendicular }}{\text { base }}$

where Ө is the angle

Substituting the values

$\left(\frac{\text {hypotenuse}}{\text {base}}\right)^{2}-\left(\frac{\text { perpendicular }}{\text {base}}\right)^{2}$

Solving,

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}$

According to Pythagoras theorem,

$\text { (hypotenuse) }^{2}-\text { (perpendicular) }^{2}=(\text { base })^{2}$

Putting this in the equation;

squared 55 - tan squared 55 =

$\frac{(\text {hypotenuse})^{2}-(\text {perpendicular})^{2}}{(\text {base}) *(\text {base})}=\frac{(\text {base})^{2}}{(\text {base}) *(\text {base})}=1$

Therefore, sec squared 55 – tan squared 55 = 1

6 0

3 years ago