Help please!!!!!!!!!!!

Mathematics

1 answer:

AysviL [449]3 years ago

7 0

Answer:

A. Plant 1 is growing at a faster rate than Plant 2.

Step-by-step explanation:

The slope (m) or rate of growth in Plant 1 is;

m = Change in y ÷ change in x

m = $\frac{3.5 - 1.5}{1 - 0}$ = 2

The linear equation of growth in Plant 1 is;

$\frac{y - 1.5}{x - 0}$ = 2

y = 2x + 1.5

The equation of growth in Plant 2 is; y = 1.5x + 3 (slope of 1.5)

Since the slope in growth of Plant 1 is greater than that in Plant 2, it implies then that Plant 1 is growing at a faster rate than Plant 2.

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Joey and Armando live on the same st as the park. The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando'

AleksandrR [38]

Let's start by assuming Armando's house is between Joey's and the park.

Let $x$ be the distance Joey walked to Armando's house.

<span>The park is 9/10 mile from Joey's home. Joey leaves home and walks to Armando's home. Then Joey and Armando walk 3/5 mile to the park.

</span> $\dfrac{9}{10} = x + \dfrac{3}{5}$

$x = \dfrac{9}{10} - \dfrac{3}{5} = \dfrac{9}{10} -\dfrac{6}{10} = \dfrac{3}{10}$

That's probably the answer they're looking for. But what if the park is between Joey and Armando's houses or Joey is between the park and Armando? (The latter isn't really possible with the given distances.)

Let $a, b, c$ be the distances between three collinear points like we have here. Our equation is really a few equations in one, something like

$\pm a \pm b = \pm c$

Let's get rid of the plus/minuses. Squaring,

$a^2 + b^2\pm 2ab = c^2$

$\pm 2ab = c^2-a^2-b^2$

$4a^2b^2 = (c^2-a^2-b^2)^2$

For us, that's a quadratic equation for $c^2$

$4(9/10)^2(3/5)^2= (c^2-(9/10)^2 - (3/5)^2)^2$

I'll skip right to the solutions,

$c^2=\dfrac{9}{100} \textrm{ or } c^2=\dfrac{9}{4}$

$c=\dfrac{3}{10} \textrm{ or } c=\dfrac{3}{2}$

We could have gotten the 3/2 just by adding 9/10+3/5 but this was more fun.

6 0

3 years ago

Some one help me pleaseeeeeeeeeee

julia-pushkina [17]

Answer:

slope = $\frac{5}{2}$