All categories

3 years ago

Gravity might be a stimulus that induces a tropic response in a plant. a. True b. False

Mathematics

1 answer:

Elena L [17]3 years ago

7 0

True. gravitropism is the positive response of the roots to grow as response to gravity.

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Let $f(x) = 2x^2 + 3x - 9,$ $g(x) = 5x + 11,$ and $h(x) = -3x^2 + 1.$ Find $f(x) - g(x) + h(x).$

Viefleur [7K]

QUESTION 1

Given that:

$f(x)=2x^2+3x-9$ ,

$g(x)=5x+11$ ,

and

$h(x)=-3x^2+1$

Then;

$f(x)-g(x)+h(x)=2x^2+3x-9-(5x+11)+(-3x^2+1)$

$f(x)-g(x)+h(x)=2x^2+3x-9-5x-11-3x^2+1$

Group similar terms;

$f(x)-g(x)+h(x)=2x^2-3x^2+3x-5x-11-9+1$

Simplify;

$f(x)-g(x)+h(x)=-x^2-2x-19$

QUESTION 2

Given that;

$f(x)=4x-7$ .

$g(x)=(x+1)^2$

and

$s(x)=f(x)+g(x)$

Substitute the functions;

$s(x)=4x-7+(x+1)^2$

Substitute x=3

$s(3)=4(3)-7+(3+1)^2$

$s(3)=12-7+(4)^2$

$s(3)=5+16$

$s(3)=21$

QUESTION 3

Given:

$f(x)=3x+2$

$g(x)=x^2-5x-1$

$f(g(x))=f(x^2-5x-1)$

This implies that;

$f(g(x))=3(x^2-5x-1)+2$

Expand the parenthesis;

$f(g(x))=3x^2-15x-3+2$

$f(g(x))=3x^2-15x-1$

QUESTION 4

The given function is;

$f(x)=3(x-6)^2+1$

Let

$y=3(x-6)^2+1$

$\Rightarrow y-1=3(x-6)^2$

$\Rightarrow \frac{y-1}{3}=(x-6)^2$

$\Rightarrow \sqrt{\frac{y-1}{3}}=x-6$

$\Rightarrow x=6+\sqrt{\frac{y-1}{3}}$

The range is:

$\frac{y-1}{3}\ge0$

$y-1\ge0$

$y\ge1$

The interval notation is;

$[1,+\infty)$

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3 years ago

How to find the minimum value of an expression

AURORKA [14]

Your answer is positive

8 0

3 years ago

Read 2 more answers

A circle has its center at (2,-4). The radius of the circle extends to the

m_a_m_a [10]

Check the picture below, so the circle looks more or less like so, with a radius of 9.

$\textit{circumference of a circle}\\\\ C=2\pi r~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=9 \end{cases}\implies C=2\pi (9)\implies C\approx 57$

3 0

2 years ago

Y = 2x + 3

mario62 [17]

The answer is no solutions

4 0

3 years ago

Use substitution to solve the following system of equations.

Diano4ka-milaya [45]

$d+e-f=-3 \\
e=f+d+7 \\
f=2e-6 \\ \\
\hbox{substitute 2e-6 for f in the 2nd equation and solve for e:} \\
e=2e-6+d+7 \\
e=2e+d+1 \ \ \ |-2e \\
-e=d+1 \ \ \ |\times (-1) \\
e=-d-1 \\ \\
\hbox{substitute -d-1 for e in the 3rd equation and solve for f:} \\
f=2(-d-1)-6 \\
f=-2d-2-6 \\
f=-2d-8$

$\hbox{substitute -d-1 for e and -2d-8 for f in the 1st equation and solve for d:} \\
d+(-d-1)-(-2d-8)=-3 \\
d-d-1+2d+8=-3 \\
2d+7=-3 \ \ \ |-7 \\
2d=-10 \ \ \ |\div 2 \\
d=-5 \\ \\ e=-d-1=-(-5)-1=5-1=4 \\ \\
f=-2d-8=-2 \times (-5)-8=10-8=2 \\ \\
\boxed{d=-5} \\ \boxed{e=4} \\ \boxed{f=2} \\
\hbox{answer D}$

5 0

3 years ago