Given rectangle QRST, what is a logical first step to prove that the diagonals QS and RT are congruent?

PLEASE HELP ME SOLVE this

Alisiya [41]

Answer:

y=-12/11

Step-by-step explanation:

y-(-12/11)=0(x-(-12/13))

y+12/11=0(x+12/13)

y+12/11=0

y=-12/11

Since the slope equals 0, then the line is horizontal.

3 0

3 years ago

Kendra ordered a set of blue and brown pins. She received 100 pins, and 30% of them were blue. How many blue pins did Kendra rec

natka813 [3]

Answer:

30 blue pins

Step-by-step explanation:

30% is the same as 30 out of 100, so this can be written in this fraction: $\frac{30}{100}$

now, you should multiply this fraction by the total number of pins she ordered:

100 x $\frac{30}{100}$

to solve this, you can simplify by dividing 100 by 100, this equals 1

30 times 1 is 30

there are 30 blue pins

8 0

2 years ago

Read 2 more answers

What are the solutions to the expression 4x2+11x−3

lina2011 [118]

Step-by-step explanation:

4x² + 11x - 3 = 0

(4x-1)(x+3) = 0

x = 1/4, -3

6 0

2 years ago

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What is the minimum value of C = 7x + 8y, given the constraints: 2x + y ≥ 8, x + y ≥ 6, x ≥ 0, y ≥ 0. A. 32 B. 42 C. 46 D. 64

marshall27 [118]

Answer: The minimum value of C is 46.

Step-by-step explanation:

Since, Here, We have to find out Min C = 7x+8y

Given the constraints are $2x+y\geq 8$ -------(1)

$x+y \geq 6$ ------------- (2)

$x \geq 0$ , $y \geq 0$ -------- (3)

Since, For equation 1) x-intercept, (4, 0) and y-intercept (0,8)

And, $2\times 0+0\geq 8$ ⇒ $0\geq 8$ ( false)

Therefore the area of line 1) does not contain the origin.

For equation 2) x-intercept, (6, 0) and y-intercept (0,6)

And, $0+0\geq 6$ ⇒ $0\geq 6$ ( false)

Therefore the area of line 2) does not contain the origin.

Thus after plotting the constraints 1) 2) and 3) we get Open Shaded feasible region AEB ( Shown in below graph)

At A≡(0,8) , C= 64

At E≡(2,4), C= 46

At B≡(6,0), C= 42

Thus at B, C is minimum, And its minimum value = 42

5 0

3 years ago

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Find the Laplace transformation of each of the following functions. In each case, specify the values of s for which the integral

MAVERICK [17]

Answer:

a. $\frac {2} {s-1}$ converges to s> 1.

b. $\frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $- \frac {2}{s + 3}$ converges to s> - 3.

d. $\frac {s}{s^2 + 25}$ converges to s> 0.

e. $\frac {10} {s^2 + 1}$ converges even s> 0.

f. $\frac {12}{s^2 + 4}$ converges to s> 0.

g. $-\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $\frac {1} {s ^ 2 + 4}$ converges to s> 0.

Step-by-step explanation:

a. $L \left\{2e^t \right\} = 2L \left\{e^t \right\} = 2 \cdot \frac {1} {s-1} = \frac {2} {s-1}$ converges to s> 1.

b. $L \left\{3e^{5t-3} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = 3e^{-3} L \left\{e^{5t} \right\} = \frac{3}{e^3 \left(s-5 \right)}$ converges to s> 5.

c. $L \left\{-2e^{-3t} \right\} = -2L \left\{e^{-3t} \right\} = - \frac {2}{s + 3}$ converges to s> - 3.

d. $L \left\{\cos\left (5t \right)\right\} = \frac {s}{s^2 + 25}$ converges to s> 0.

e. $L \left\{10 \sin\left(t\right)\right\} = 10L\left\{\sin\left(t\right)\right\} = \frac {10} {s^2 + 1}$ converges even s> 0.

f. $L \left\{6\sin \left(2t \right) \right\} = 6L\left\{\sin\left (2t\right)\right\} = \frac {12}{s^2 + 4}$ converges to s> 0.

g. $L \left\{-5\cos\left(2t + 1\right) \right\} = -5L\left\{\cos\left(2t + 1 \right)\right\} = -\frac {5\left(\cos\left (1\right) s-2 \sin\left(1\right)\right)}{s^2 + 4}$ converges to s> 0.

h. $L\left\{\sin \left(t\right)\cos \left(t\right)\right\} = L\left\{\sin\left(2t\right)\frac{1}{2}\right\} =\frac{1}{2}\cdot \frac{2}{s^2+4} = \frac {1} {s ^ 2 + 4}$ converges to s> 0.

7 0

3 years ago