Wich systems of equations would have NO solution? (Answer with work for branliest)

Mathematics

1 answer:

Ierofanga [76]3 years ago

5 0

for 1st

comparing value

3x+2=-3x-2

6x=-4

x=-4/6Have solution.

for 2nd.

y=-2x+10

both are equal So no solution.

3rd

y=-x+4

both are equal So no solution

4th

3x+y=8

3x+y=9

both are equal So no solution

You might be interested in

Solve the following quadratic equation using the quadratic formula. Which of the following expressions gives the numerators of t

BabaBlast [244]

I think it would be the third one

6 0

4 years ago

B. 8/27 D. 1/3 10. Three spheres of radii , 2r, and 3r, respectively, are melted and formed into a new sphere. Find the surface

Dahasolnce [82]

Answer:

$A_{f}=4\pi (\sqrt[3]{36} r)^{2}\\\\V_{f}=\frac{4}{3} \pi (36r^{3})$

Step-by-step explanation:

In order to find the final radii of the sphere, we need to calculate the volume, knowing that volumes are additive:

$V_{1}=\frac{4}{3} \pi (r^{3})\\\\V_{2}=\frac{4}{3} \pi (2r)^{3}\\\\V_{3}=\frac{4}{3} \pi (3r)^{3}\\\\V_{f}=\frac{4}{3} \pi (r^{3}+(2r)^{3}+(3r)^{3})\\\\V_{f}=\frac{4}{3} \pi (r^{3}+8r^{3}+27r^{3})\\\\V_{f}=\frac{4}{3} \pi (36r^{3})\\\\V_{f}=\frac{4}{3} \pi R^{3}\\\\R=\sqrt[3]{36} r$