All categories

3 years ago

HELP ME AND ILL GIVE U BRAINLIST I PROMISE BUT ITS ONLY IF U GET IT RIGHT

Mathematics

2 answers:

ivanzaharov [21]3 years ago

6 0

Answer:

Step-by-step explanation:

emmainna [20.7K]3 years ago

3 0

Answer:

if it says she used 4 cups of sugar then she used 4 cups of sugar?

but if she used 4 cups of flour then she used 2 cups of sugar

Step-by-step explanation:

Hope this helps! :)

You might be interested in

1) 8(4w - 5) = 2) 7 (-35 - 6) - can you help me solve these

tankabanditka [31]

1) 32w -40
2) -275

Lmk if it helped you

7 0

3 years ago

Read 2 more answers

The table shows the relationship y = kx.

sammy [17]

Answer: it’s 1/5

Step-by-step explanation:

6 0

3 years ago

Please answer this correctly which is the correct options

MakcuM [25]

The first two are equivalent to 15 = k + 13

8 = k + 13 - 7 → ( add 7 to both sides )

8 + 7 = k + 13 ⇒ 15 = k + 13 ' is equivalent '

12 = k + 13 - 3 → ( add 3 to both sides )

12 + 3 = k + 13 ' is equivalent '

4 0

3 years ago

Verify cos^4x-sin^4x=cos2x

kati45 [8]

A^2 - b^2 = (a - b)( a +b);
cos^4x-sin^4x =( cos^2x + sin^2x )( cos^2x - sin^2x) = 1* cos2x = co2x.

7 0

3 years ago

Using the method of completing the square, put each circle into the form

tatiyna

Answer:

Standard form: $(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Center: $(\frac{1}{2}, 0)$

Radius: $r =\sqrt{15}$

Step-by-step explanation:

The equation of a circle in the standard form is

$(x-h)^{2} + (y-k)^{2} = r^{2}$

Where the point (h, k) is the center of the circle

To transform this equation $4x^{2} -4x + 4y^{2} - 59 = 0$ this equation in the standard form we use the method of square.

First, we group similar variables

$(4x^{2} -4x) + (4y^{2}) - 59 = 0$

Divide both sides of equality by 4

$(x^{2} -x) + (y^{2}) - 14.75 = 0$

Now we complete square for variable x.

Take the coefficient "b" that accompanies the variable x and divide by 2. Then, elevate the result to the square:

$b =-1\\\\\frac{b}{2}= \frac{-1}{2}= -\frac{1}{2}\\\\(\frac{b}{2})^2= (-\frac{1}{2})^2 = \frac{1}{4}$

Now add $(\frac{b}{2})^2$ on both sides of the equality

$(x^{2} -x +\frac{1}{4}) + (y^{2}) - 14.75 = (\frac{1}{4})$

Factor the expression and simplify the independent terms

$(x-\frac{1}{2})^2 + (y^{2}) = 15$

$(x-\frac{1}{2})^2 + (y-0)^2 = 15$

Then

$h =\frac{1}{2}\\\\k=0$

and the center is $(\frac{1}{2}, 0)$

radius $r =\sqrt{15}$

3 0

3 years ago