All categories

3 years ago

Of a gallon of jambalaya. If Jack's recipe is for 1 gallon of jambalaya, how much okra will he

Mathematics

1 answer:

serg [7]3 years ago

8 0

Answer:

2 bags of okra

Step-by-step explanation:

Double means 2 and 1 times 2 is equal to 2. Hope this helps

You might be interested in

Priscilla’s grandmother’s fruit salad recipe calls for one part apple, one part orange, four parts strawberry, two parts cherry,

Anvisha [2.4K]

Answer:

Step-by-step explanation:

There are 11 parts of fruit: 4-strawberry, 3- grape, 2, cherry, 1- apple, 1-orange

4/11 from the entine quantity of fruits- strawberry

3/11-grapes

2/11- cherries

1/11- apples

1/11- oranges } ⇒ the quantity of oranges and apples are equal

The biggest part of the fruit- 4/11 - are strawberries

4/11>3/11>2/11>1/11

the smallest part of the fruit- apples and oranges

5 0

3 years ago

Leigh took out a payday loan for $400 due in 2 weeks that charged a $50 fee. What is the periodic interest rate of the loan?

VMariaS [17]

Answer: I got 12.5 for apex

Step-by-step explanation:

I just took the test

6 0

3 years ago

Read 2 more answers

A hobbyist is making a toy sailboat. For the triangular sail, she wants the height h (in inches)

posledela

Answer:

Step-by-step explanation:

7 0

3 years ago

What is this I need help don’t mind the room

beks73 [17]

Here’s a picture of my work. The answer is b. Hope this helps :-)

3 0

2 years ago

Let alpha and beta be conjugate complex numbers such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}

miskamm [114]

Answer:

$-3+i\sqrt{3} , 1+\sqrt{3}$

Step-by-step explanation:

Given that alpha and beta be conjugate complex numbers

such that frac{\alpha}{\beta^2} is a real number and alpha - \beta| = 2 \sqrt{3}.

Let

$\alpha = x+iy\\\beta = x-iy$

since they are conjugates

$\alpha-\beta = x+iy-(x-iy)\\= 2iy= 2i\sqrt{3} \\y =\sqrt{3}$

$\frac{\alpha}{\beta^2} }\\=\frac{x+i\sqrt{3} }{(x-i\sqrt{3})^2} \\=\frac{x+i\sqrt{3}}{x^2-3-2i\sqrt{3}} \\=\frac{x+i\sqrt{3}((x^2-3+2i\sqrt{3}) }{(x^2-3-2i\sqrt{3)}(x^2-3-2i\sqrt{3})}$

Imaginary part of the above =0

i.e. $\sqrt{3} (x^2-3)+2x\sqrt{3} =0\\x^2+2x-3=0\\(x+3)(x-1) =0\\x=-3,1$

So the value of alpha = $-3+i\sqrt{3} , 1+\sqrt{3}$

3 0

3 years ago