Find the value of t for the given values of d and r. d = 286, r = 55

Please help and thank you

tatiyna

Answer:

$b_{1}=\frac{2A}{h}-b_{2}$

Step-by-step explanation:

Given: $A=\frac{1}{2}(b_{1} +b_{2})h$

We need to completely isolate $b_{1}$ to solve.

$A=\frac{1}{2}(b_{1} +b_{2})h$

$A=(\frac{1}{2}b_{1} +\frac{1}{2} b_{2})h$

$A=\frac{1}{2}b_{1} h+\frac{1}{2}b_{2}h$

$-\frac{1}{2}b_{1}h+A=\frac{1}{2}b_{2}h$

$-\frac{1}{2}b_{1}h=-A+\frac{1}{2}b_{2}h$

$-\frac{1}{2}b_{1}=\frac{-A}{h}+\frac{1}{2}b_{2}$

Finally, multiply both sides by -2 to completely isolate $b_{1}$ .

$b_{1}=\frac{2A}{h}-b_{2}$

5 0

2 years ago

Read 2 more answers

The price of a laptop is $420. The price of a phone is 35% of the price of this laptop. What is the price of the phone? A $12 B

Pavel [41]

The correct answer would be B. $420x0.35=$147.00

8 0

2 years ago

Read 2 more answers

Write an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the line described by y=1/2

Marat540 [252]

Given:

The point, (4, -3)

The line,

$y=\frac{1}{2}x+5$

To find an equation in slope-intercept form for the line that passes through (4,-3) and is parallel to the given line:

The slope of the line is,

$m=\frac{1}{2}$

Since the given line is parallel to the new line, so the slope will be same for the both.

Using the point-slope formula,

$y-y_1=m(x-x_1)$

Substitute the point and slope we get,

$\begin{gathered} y-(-3)=\frac{1}{2}(x-4) \\ y+3=\frac{1}{2}x-2 \\ y=\frac{1}{2}x-2-3 \\ y=\frac{1}{2}x-5 \end{gathered}$

Hence, the equation in slope-intercept form for the line is,

$y=\frac{1}{2}x-5$

8 0

1 year ago

In each case, determine if the function can be differentiated using the rules developed so far (sections 3.1 and 3.2). (a) y = 2

Alex_Xolod [135]

The rule for differentiation for variable with exponents has the formula:

d/dx (xⁿ) = nxⁿ⁻¹
where n is the exponent

Thus, for the given equation, the solution is as follows:
y = 2x²
dy/dx = 2(2)x²⁻¹ = 4x

Thus, the derived equation of the given would now be 4x.

3 0

2 years ago

Triangle TUV was dilated to create triangle T'U'V' using point A as the center of dilation. What is the scale factor of the dila

Tomtit [17]

A dilation is a transformation, with center O and a scale factor of k that is not zero, that maps O to itself and any other point P to P'. The center O is a fixed point, P' is the image of P, points O, P and P' are on the same line.

Thus, a dilation with centre O and a scale factor of k maps the original figure to the image in such a way that the distances from O to the vertices of the image are k times the distances from O to the original figure. Also the size of the image are k times the size of the original figure.

In the dilation of triangle TUV, It is obvious that the image T'U'V' is smaller than the original triangle TUV and hence the scale factor is less than 1.

The ratio of the distances from A to the vertices of the image T'U'V' to the distances from A to the original triangle TUV is the scale factor.

The scale factor = 3.2 / 4.8 = 2/3

8 0

2 years ago

Read 2 more answers