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3 years ago

In order to isolate the variable term in the equation 6 - 2x = -3 using the subtraction property of equality, which number

Mathematics

1 answer:

kvv77 [185]3 years ago

6 0

Answer:

x=9/2

Step-by-step explanation:

6-2x=-3

2x=6-(-3)

2x=6+3

2x=9

x=9/2

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The school band is ordering health bars to sell for a fundraiser. The company that sells the bars charges $0.40 per bar plus $20

Illusion [34]

0-449 is the most reasonable bc of how much the shipping costs

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2 years ago

Janae has 280 dimes in her piggy bank she has 10 times as many pennies how many pennies does Janae have

Natalija [7]

Answer:

She has 28 pennies

Step-by-step explanation:

Janae has 280 dimes in her piggy bank she has 10 times as many pennies .

If she has 280 dimes in piggy bank

And has 10 times as many pennies

Let dimes= x

Let pennies= y

X= 10y

But x= 280

280= 10y

280/10 = y

28 = y

She has 28 pennies

8 0

3 years ago

How dou you solve for 6x+4y=24?

lilavasa [31]

You could first solve for y, assuming x is 0, and y = 6. Then solve for x, and that means y = 4.

3 0

3 years ago

(Show your working) I uploaded the question

svlad2 [7]

x^2 = the first integer

(x - 1)^2 = the second integer.

x^2 - (x - 1)^2 = ?

First, let's plug a number into our equation for x.

(2)^2 - (2 - 1)^2 = ?

4 - (1)^2 = ?

4 - 1 = 3

As we can see the difference is odd but it's also the sum of the two consecutive integers.

2 + 1 = 3.

This works for all numbers. Let's plug another number into our equation for x.

(4)^2 - (4 - 1)^2 = ?

16 - (3)^2 = ?

16 - 9 = 7

4 + 3 = 7

Try any number and it will always be odd.

4 0

3 years ago

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