All categories

2 years ago

Answer as much as youd like, and pls pls pls dont take my points I actually need help

Mathematics

2 answers:

crimeas [40]2 years ago

6 0

Answer:

We don't take them we earn them when we solve your question

Step-by-step explanation:

Talja [164]2 years ago

5 0

Exactly, because you guys put out the answers and we get points because we answer your guys questions when you need help.
Step by Step:

You might be interested in

Sierra is growing tomato plants. She wants to grow the plants to be at least 48 inches tall. Which situation is represented by t

Len [333]

The first one is the correct answer due to 4 is the initial height of the plaint and 0.75 are inches that being added every week

6 0

3 years ago

Read 2 more answers

Can someone help me please !

elena55 [62]

The reference angle is
pi/3.

At 180 it would be 9pi/3 so going past that is 10pi/3.

7 0

2 years ago

A drawer contains 2 red shirts and 4 blue shirts. A second drawer contains 3 pairs of grey pants and 2 pairs of blue pants. A th

viktelen [127]

Answer:

b . may be

may it helped u ....

4 0

2 years ago

Write the explicit formula that represents the geometric sequence -2, 8, -32, 128

lora16 [44]

$a_{1}=-2 \\ \\ a_{2}=8 \\ \\ q= \frac{a_{2}}{a_{1}} \\ \\ q= \frac{8}{-2}=-4 \\ \\ a_{n}=a_{1}*q^{n-1} \\ \\ a_{n}=-2*-4^{n-1}$

7 0

2 years ago

f(x) is a quadratic function with x-intercepts at (−1, 0) and (−3, 0). If the range of f(x) is [−4, [infinity]) and g(x) = 2x^2

iren2701 [21]

Answer with Step-by-step explanation:

We are given that function f(x) which is quadratic function.

x -intercept of function f(x) at (-1,0) and (-3,0)

x-Intercept of f means zeroes of f

x=-1 and x=-3

Range of f =[-4, $\infty$ )

g(x)= $2x^2+8x+6=2(x^2+4x+3)$

$g(x)=0$

$2(x^2+4x+3)=0$

$x^2+4x+3=0$

$x^2+3x+x+3=0$

$x(x+3)+1(x+3)=0$

$(x+1)(x+3)=0$

$x+1=0\implies x=-1$

$x+3=0\implies x=-3$

Therefore, x-intercept of g(x) at (-1,0) and (-3,0).

Substitute x=-2

$g(-2)=2(-2)^2+8(-2)+6=8-16+6=-2$

$g(x)=2(x^2+4x)+6$

$g(x)=2(x^2+2\times x\times 2+4-4)+6=2(x^2+2\times x\times 2+4)-8+6$

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

By comparing with the equation of parabola

$y=a(x-h)^2+k$

Where vertex=(h,k)

We get vertex of g(x)=(-2,-2)

Range of g(x)=[-2, $\infty$ )

Zeroes of f and g are same .

But range of f and g are different.

Range of f contains -3 and -4 but range of g does not contain -3 and -4.

f and g are both quadratic functions.

3 0

2 years ago