All categories

2 years ago

Solve the following system of equations: {−6x−y=2−7x−4y=−9

Mathematics

1 answer:

Nata [24]2 years ago

6 0

Answer:

x=-1, y=4. (-1, 4).

Step-by-step explanation:

-6x-y=2

-7x-4y=-9

----------------

y=-6x-2

-7x-4(-6x-2)=-9

-7x+24x+8=-9

17x=-9-8

17x=-17

x=-17/17

x=-1

-6(-1)-y=2

6-y=2

y=6-2=4

You might be interested in

I NEED HELP ASAP PLEASE!

dusya [7]

Answer:

The length of AC = 4 and the length of A'C' = 1 so the scale factor is 1/4.

8 0

3 years ago

hi i'm really struggling with this and its due later today! If someone could show the work of how to do it and the answer that w

GuDViN [60]

Answer:

$\frac{-5 \pm \sqrt{13} }{6}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Step-by-step explanation:

Step 1: Define

$\frac{-5 \pm \sqrt{5^2-4(3)(1)} }{2(3)}$

Step 2: Evaluate

Evaluate Exponents: $\frac{-5 \pm \sqrt{25-4(3)(1)} }{2(3)}$
Evaluate Multiplication: $\frac{-5 \pm \sqrt{25-12} }{6}$
Evaluate Subtraction: $\frac{-5 \pm \sqrt{13} }{6}$

8 0

3 years ago

Help please! (no one was answering it in physics and i really need help)

aivan3 [116]

Let $v_0$ be the cat's speed just as it leaves the edge of the table. Then taking the point 1.3 m below the edge of the table to be the origin, the cat's horizontal position at time $t$ is given by

$x(t)=v_0t$

and its height is

$y(t)=1.3\,\mathrm m-gt^2$

where $g$ is 9.8 m/s^2, the magnitude of the acceleration due to gravity.

The time it takes for the cat to hit the ground is $t$ with

$0=1.3\,\mathrm m-gt^2\implies t=\sqrt{\dfrac{1.3\,\rm m}g}\approx0.36\,\mathrm s$

(Unfortunately, this doesn't match any of the given options...)

The cat lands 0.75 m away (horizontally) from the edge of the table, so that its speed $v_0$ was

$0.75\,\mathrm m=v_0(0.36\,\mathrm s)\implies v_0\approx2.08\dfrac{\rm m}{\rm s}$

(Again, not one of the answer choices...)

I'm guessing there's either a typo in the question or answers.

6 0

3 years ago

a square sandbox has an area of 42 squareft what is the approximate perimeter of the sandbox to the nearest hundredth

Ugo [173]

The area of a square is its side length squared. So the side length of the square squared is 42, so the side length of the square is sqrt(42). The perimeter of the square is 4 times the side length, or 4sqrt(42), which is approximately 25.92.

3 0

3 years ago

Write the equation of the circle graphed below.

mojhsa [17]

Answer:

As shown in picture, this circle has radius 1.5 and passes (0, 1.5)

=> According to the general form of equation of circle that has radius r and passes (a, b): (x - a)^2 + (y - b)^2 = r^2, we have:

x^2 + (y - 1.5)^2 = 1.5^2

<=>

x^2 + (y - 1.5)^2 = 2.25

Hope this helps!

6 0

3 years ago