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3 years ago

5

A landscape contractor planted 37 tulip bulbs, 53 daffodil bulbs, and 82 crocus bulbs in each field. He planted bulbs in 174 fie

lds. What is the total number of bulbs the landscape contractor planted?
A.
172

B.
346

C.
28,188

D.
29,928

1 answer:

ohaa [14]3 years ago

3 0

He planted D. 29,928 Bulbs
First Add up all the bulbs.
Then multiply the # of bulbs with the # of fields.

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Leon receives a birthday gift of $50 from his parents. Each week he saves $5.00. How many weeks will it take for him to save $12

lubasha [3.4K]

Answer:

14 weeks. the equation would be a.

Step-by-step explanation:

120-50=70 70/5=14 it will take him 14 weeks to get 120 if he saves 5 a week on top of the additional 50 he already has.

3 0

3 years ago

What 2 numbers add to get 2 and multiply to 4

rusak2 [61]

Factor 4
4=1 times 4
2 times 2
they don't add to 2
set up equation
x+y=2
xy=4

first equation, subtract x from both sides
y=2-x
subsitute for y
x(2-x)=4
distribute
2x-x^2=4
add x^2
2x=x^2+4
subtract 2x
0=x^2-2x+4
use quadratic formula which is
if you have ax^2+bx+c=0 then
x= $\frac{ -b+/-\sqrt{b^{2}-4ac} }{2a}$ so

1x^2-2x+4=0
a=1
b=-2
c=4
x= $\frac{ -(-2)+/-\sqrt{(-2)^{2}-4(1)(4)} }{2(1)}$
x= $\frac{ 2+/-\sqrt{4-16} }{2}$
x= $\frac{ 2+/-\sqrt{-12} }{2}$
we have $\sqrt{-12}$ and that doesn't give a real solution
therefor there are no real solutions
but if you want to solve fully
x= $\frac{ 2+/-2\sqrt{-3} }{2}$
i= $\sqrt{-1}$
x= $\frac{ 2+/-2i\sqrt{3} }{2}$
x= $1+/-i\sqrt{3}$
x= $1-i\sqrt{3}$ or x= $1+i\sqrt{3}$ (those are the 2 numbers)

6 0

4 years ago

Find the linear approximation of the function g(x) = 3 root 1 + x at a = 0. g(x). Use it to approximate the numbers 3 root 0.95

Virty [35]

Answer:

$L(x)=1+\dfrac{1}{3}x$

$\sqrt[3]{0.95} \approx 0.9833$

$\sqrt[3]{1.1} \approx 1.0333$

Step-by-step explanation:

Given the function: $g(x)=\sqrt[3]{1+x}$

We are to determine the linear approximation of the function g(x) at a = 0.

Linear Approximating Polynomial, $L(x)=f(a)+f'(a)(x-a)$

a=0

$g(0)=\sqrt[3]{1+0}=1$

$g'(x)=\frac{1}{3}(1+x)^{-2/3} \\g'(0)=\frac{1}{3}(1+0)^{-2/3}=\frac{1}{3}$

Therefore:

$L(x)=1+\frac{1}{3}(x-0)\\\\$The linear approximating polynomial of g(x) is:$\\\\L(x)=1+\dfrac{1}{3}x$

(b) $\sqrt[3]{0.95}= \sqrt[3]{1-0.05}$

When x = - 0.05

$L(-0.05)=1+\dfrac{1}{3}(-0.05)=0.9833$

$\sqrt[3]{0.95} \approx 0.9833$

(c)

(b) $\sqrt[3]{1.1}= \sqrt[3]{1+0.1}$

When x = 0.1

$L(1.1)=1+\dfrac{1}{3}(0.1)=1.0333$

$\sqrt[3]{1.1} \approx 1.0333$

7 0

3 years ago

The problem is in the picture

creativ13 [48]

It should be the first answer choice between those two

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3 years ago

CHECK YOUR KNOWLEDGE

PIT_PIT [208]

1. A rent paid
2. B 10

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4 years ago