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3 years ago

Evaluate the expression for a = 30, b = 30, c = 7, d = 24: a + b = c = d.

Mathematics

1 answer:

suter [353]3 years ago

4 0

Answer:

Not possible

Step-by-step explanation:

a + b = c = d Substitute

30 + 30 = 7 = 24 Add

60 = 7 = 24 This is not possible because these numbers aren't equal.

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2y-2z= -4 - 4y+2z = -2 x+4y-z=-5

Bumek [7]

Answer:

(-12,3,5)

x=-12,y=3.z=5

Step-by-step explanation:

6 0

3 years ago

Graph the linear inequality. 3x – 4y > 8 HELP ASAP

DedPeter [7]

Brainly isn’t letting me insert a picture but the anwser is y<2+3x/4

7 0

3 years ago

3 over 12 in simplest form

Novosadov [1.4K]

Answer:

1/4

Step-by-step explanation:

3/12

Both the top and bottom number can be divided by 3

3 / 3

12 / 3

1/ 4

7 0

4 years ago

Use substitution to solve the system of equations 4x+y=0 2y+x=-7

RSB [31]

Move the 1st equation to get y by itself so it’s now y=-4x so you can substitute that in for the second equation 2y+x=-7 becomes 2(-4x)+x=-7
-8x+x=-7
-7x=-7
x=1

then plug that back in to find y
4(1)+y=0
4+y=0
y=-4

now you have x and y so your coordinate (the answer) is (1, -4) hope this helps!

5 0

4 years ago

The hypotenuse of a right triangle is 6 cm and one side is 2 cm longer than the other side. Find the length of each side to one

MA_775_DIABLO [31]

Answer:

The two legs of the right triangle are approximately 3.1 cm and 5.1 cm and the hypotenuse is 6 cm.

Step-by-step explanation:

Let x be one leg of the right triangle.

Since the other leg is two centimeters longer, it can be represented by the expression:

$x + 2 \text{ cm}$

According to the Pythagorean Theorem, for a right triangle:

$a^2 + b^2 = c^2$

Where c is the hypotenuse and a and b are the two legs.

The hypotenuse is given to be 6 cm and the two legs are x and (x + 2). Hence:

$(x)^2 + (x+2)^2 = 6^2$

Solve for x. Simplify:

$x^2 + (x^2 + 4x + 4) = 36$

Simplify:

$2x^2 + 4x + 4 = 36$

Subtract:

$2x^2 + 4x - 32 =0$

Divide:

$x^2 + 2x - 16 = 0$

The equation is not factorable, so we can consider using the quadratic formula:

$\displaystyle x = \frac{-b\pm\sqrt{b^2 -4ac}}{2a}$

In this case, a = 1, b = 2, and c = -16. Substitute:

$\displaystyle x= \frac{-(2)\pm\sqrt{(2)^2-4(1)(-16)}}{2(1)}$

And evaluate:

$\displaystyle \begin{aligned}x &= \frac{-2\pm\sqrt{68}}{2} \\ \\ &= \frac{-2\pm\sqrt{4\cdot 17}}{2} \\ \\ &= \frac{-2\pm2\sqrt{17}}{2} \\ \\ &= -1 \pm \sqrt{17} \end{aligned}$

Hence, our two solutions are:

$\displaystyle x = -1 + \sqrt{17} \approx 3.1\text{ or } x = -1 - \sqrt{17} \approx -5.1$

Lengths cannot be negative, so we can ignore the second solution.

Hence, the value of x or the first side length is about 3.1 centimeters.

Since the other side length is two centimeters longer, the other side is about 5.1 centimeters.

In conclusion, the two legs of the right triangle are approximately 3.1 cm and 5.1 cm and the hypotenuse is 6 cm.

4 0

3 years ago