All categories

3 years ago

I just need the answer thanks. D is 57 ouces if you can't see it

Mathematics

2 answers:

UkoKoshka [18]3 years ago

8 0

Answer:

c) 20 ounces

Step-by-step explanation:

jeka943 years ago

7 0

C. 20 ounces__________

You might be interested in

£280 is divided between Krutika, Natasha & Richard so that Krutika gets twice as much as Natasha, and Natasha gets three tim

Anna11 [10]

300 nata sell because of your new job and I don’t know

7 0

2 years ago

Pls help giving brbrainlyist

chubhunter [2.5K]

The fourth choice. For example, (x+2)(x-2)>0 we'll get x>2 or x<-2 because whenever you substitute x>2 for example x = 3, the value will be higher than 0 itself.

If we substitute x = 3 (for x>2)

We'll get (3+2)(3-2)>0

5(1)>0

5>0 The statement is true.

If we substitute x = -3 (for x<-2)

We'll get (-3+2)(-3-2)>0

-1(-5)>0

5>0 The statement is true.

We don't say x>2 AND x<-2, In math, we only say x>2 OR x<-2

6 0

3 years ago

Read 2 more answers

When it is 2 hours after 2 o'clock, then it is 4 o'clock (2 + 2 = 4). When it is 10 hours after 10 o'clock, then it is 8 o'clock

lbvjy [14]

It will be 4 o'clock when Brad takes his 16th pill.

Hope it helps

4 0

3 years ago

Read 2 more answers

Find all points on the x-axis that are 14 units from the point (6,-7) All points on the x-axis that are 14 units from the point

Maksim231197 [3]

Answer: $(6+7\sqrt{3},0)\text{ and }(6-7\sqrt{3},0)$ are the required points.

or (18.124,0) and ( -6.124,0) are the required points.

Step-by-step explanation:

Let (x,0) be the point on x -axis that are 14 units from the point (6,-7) .

Then by distance formula , we have

$\sqrt{(x-6)^2+(0-(-7))^2}=14\ \ \ [\ \because distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}]$

Taking square on both the sides , we get

$(x-6)^2+7^2=14^2\\\\\Rightarrow\ x^2+6^2-2(6)x+49=196\\\\\Rightarrow\ x^2+36-12x=147\\\\\Rightarrow\ x^2-12x=111\\\\\Rightarrow\ x^2-12x-111=0$

Using quadratic formula : $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x=\dfrac{12\pm\sqrt{(-12)^2-4(1)(-111)}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{144+444}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{588}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm\sqrt{2^2\times7^2\times3}}{2}\\\\\Rightarrow\ x=\dfrac{12\pm14\sqrt{3}}{2}\\\\\Rightarrow\ x=6\pm7\sqrt{3}$