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lapo4ka [179]
2 years ago
6

Rowena used small wooden cubes to measure the length of her foot and her hand. Her foot was 22 cubes long and her had was 15 cub

es long. Which number sentence should Rowena use to find the difference between these lengths?
A. 22 + 15
B. 22-15
C. 22 x 15
D. 22 ÷ 15
Mathematics
2 answers:
Nitella [24]2 years ago
8 0
The answer is B, because it is asking for the difference which is subtraction.
Sonja [21]2 years ago
5 0
A. Or B.

I picked this in one of my questions and I don’t remember which specific one but the one or other should be correct.
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Advocard [28]
I think the answer to your question is 15 hope this helps
6 0
3 years ago
What is the equation of the line that passes through the point (5,6) and has a slope of 2 ?
Serhud [2]

Answer:

y = 2x - 4

Step-by-step explanation:

The equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y- intercept )

Here m = 2, thus

y = 2x + c ← is the partial equation

To find c substitute (5, 6) into the partial equation

6 = 10 + c ⇒ c = 6 - 10 = - 4

y = 2x - 4 ← equation of line

4 0
2 years ago
A motorboat travels Downstream on a river for 5 hours. The speed of the current is 2 km/h. The river flows into a still Lake and
lara [203]

Answer:

I think it is 10 km/hr

Step-by-step explanation:

5 hrs = 40 more with 2 km/hr current

2 hrs= less 40 no current

40-2x5=30

30 without current for three hours

30/3=10 km/hr no current

CHECK

10x2=20

10x5=50+2x5=60

8 0
2 years ago
Rewrite the equation by completing the square.2xE2+7x+6=0
kodGreya [7K]

Answer:

(X+7/4)=1/16

Step-by-step explanation:

.

8 0
3 years ago
find the coordinates of the point P on the parabola y=1-x^2 with domain 0≤x≤1 that minimize the area of the triangle enclosed by
coldgirl [10]

Let point P be with coordinates (x_0,y_0). Find the equation of the  tangent line.

1. If y=1-x^2, then y'=-2x.

2. The equation of the tangent line at point P is

y-y_0=-2x_0(x-x_0).

Find x-intercept and y-intercept of this line:

  • when x=0, then y=y_0+2x_0^2;
  • when y=0, then x=\dfrac{y_0}{2x_0}+x_0=\dfrac{y_0+2x_0^2}{2x_0}.

The area of the triangle enclosed by the tangent line at P, the x-axis, and y-axis is

A=\dfrac{1}{2}\cdot (2x_0^2+y_0)\cdot \left(\dfrac{y_0+2x_0^2}{2x_0}\right)=\dfrac{(y_0+2x_0^2)^2}{4x_0}.

Since point P is on the parabola, then y_0=1-x_0^2 and

A=\dfrac{(1-x_0^2+2x_0^2)^2}{4x_0}=\dfrac{(1+x_0^2)^2}{4x_0}.

Find the derivative A':

A'=\dfrac{2(1+x_0^2)\cdot 2x_0\cdot 4x_0-4(1+x_0^2)^2}{16x_0^2}=\dfrac{12x_0^4+8x_0^2-4}{16x_0^2}.

Equate this derivative to 0, then

12x_0^4+8x_0^2-4=0,\\ \\3x_0^4+2x_0^2-1=0,\\ \\D=2^2-4\cdot 3\cdot (-1)=16,\ \sqrt{D}=4,\\ \\x_0^2_{1,2}=\dfrac{-2\pm4}{6}=-1,\dfrac{1}{3},\\ \\x_0^2=\dfrac{1}{3}\Rightarrow x_0_{1,2}=\pm\dfrac{1}{\sqrt{3}}.

And

y_0=1-\left(\pm\dfrac{1}{\sqrt{3}}\right)^2=\dfrac{2}{3}.

Answer: two points: P_1\left(-\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right), P_2\left(\dfrac{1}{\sqrt{3}},\dfrac{2}{3}\right).

6 0
2 years ago
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