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3 years ago

If you can buy 6 bulbs of elephant garlic for $18 then how many can you buy with $54?

Mathematics

2 answers:

AlexFokin [52]3 years ago

4 0

Answer:

18 bulbs for $54

Step-by-step explanation:

if we get 6 bulbs for $18

we can keep going

12 bulbs for $36

18 bulbs for $54

or we can divide 54 by 18 and get 3

then we multiply 3 by 6 and get 18 bulbs :)

if im correct may i please get brainliest

emmasim [6.3K]3 years ago

4 0

Answer:

18 bulbs

Step-by-step explanation:

let x number of bulbs can be bought for $54.

$\frac{6}{18} =\frac{x}{54} \\\frac{1}{3} =\frac{x}{54} \\\\3x = 54\\x =\frac{54}{3} \\x=18$

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Pls help a girl out .......

VARVARA [1.3K]

Answer:

Yes

Step-by-step explanation:

The sentence does correctly state the equation written.

8 0

3 years ago

The endpoints of RS are R(–5, 12) and S(4, –6). What are the coordinates of point T, which divides RS into a 4:5 ratio?

Ksivusya [100]

Answer:

Option 2 is correct that is (-1,4)

Step-by-step explanation:

We will use the section formula to find the coordinates of T

Formula is $(x,y)=\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}$

Here, $m_1=4,m_2=5,x_1=-5,y_1=12,x_2=4,y_2=-6$

On substituting the values in the formula we get

$(x,y)=\frac{4\cdot 4+5\cdot -5}{4+5},\frac{4\cdot -6+5\cdot 12}{4+5}$

$\Rightarrow (x,y)=\frac{4\cdot 4+5\cdot -5}{4+5},\frac{4\cdot -6+5\cdot 12}{4+5}$

$\Rightarrow (x,y)=\frac{16-25}{9},\frac{-24+60}{9}$

$\Rightarrow (x,y)=\frac{-9}{9},\frac{36}{9}$

$\Rightarrow (x,y)=(-1,4)$

Therefore, option 2 is correct that is (-1,4)

4 0

3 years ago

Read 2 more answers

A quadrilateral has three angles that measure 90o , 100o, and 120o. Which is the measure of the fourth angle?

Reptile [31]

$360 - (100 + 90 + 120) \\ 360 - 310 = 50$
it measures 50

good luck

3 0

3 years ago

4. Using the geometric sum formulas, evaluate each of the following sums and express your answer in Cartesian form.

nikitadnepr [17]

Answer:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=1$

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=0$

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})=\frac{1}{2}$

Step-by-step explanation:

$\sum_{n=0}^9cos(\frac{\pi n}{2})=\frac{1}{2}(\sum_{n=0}^9 (e^{\frac{i\pi n}{2}}+ e^{\frac{i\pi n}{2}}))$

$=\frac{1}{2}(\frac{1-e^{\frac{10i\pi}{2}}}{1-e^{\frac{i\pi}{2}}}+\frac{1-e^{-\frac{10i\pi}{2}}}{1-e^{-\frac{i\pi}{2}}})$

$=\frac{1}{2}(\frac{1+1}{1-i}+\frac{1+1}{1+i})=1$

2nd

$\sum_{k=0}^{N-1}e^{\frac{i2\pi kk}{2}}=\frac{1-e^{\frac{i2\pi N}{N}}}{1-e^{\frac{i2\pi}{N}}}$

$=\frac{1-1}{1-e^{\frac{i2\pi}{N}}}=0$

3th

$\sum_{n=0}^\infty (\frac{1}{2})^n cos(\frac{\pi n}{2})==\frac{1}{2}(\sum_{n=0}^\infty ((\frac{e^{\frac{i\pi n}{2}}}{2})^n+ (\frac{e^{-\frac{i\pi n}{2}}}{2})^n))$

$=\frac{1}{2}(\frac{1-0}{1-i}+\frac{1-0}{1+i})=\frac{1}{2}$

What we use?

We use that

$e^{i\pi n}=cos(\pi n)+i sin(\pi n)$

and

$\sum_{n=0}^k r^k=\frac{1-r^{k+1}}{1-r}$

6 0

3 years ago

What’s the answer?? Pls help

Artemon [7]

Answer:

Step-by-step explanation:

8 0

3 years ago