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2 years ago

Find the greatest common factor of 6m² and 7m².

Mathematics

1 answer:

77julia77 [94]2 years ago

8 0

Answer:

your answer is 4 my friend

Step-by-step explanation:

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Use the table to identify the value of p and q that should be used to factor x^2-3x-10 as (x+p)(x+q) I’ve seen the answer with p

Elza [17]

Answer:

Step-by-step explanation:

1 is the right answer for this question

3 0

3 years ago

Evaluate the expression without using a calculator Tan(arccos3/4)

julia-pushkina [17]

9514 1404 393

Answer:

(√7)/3

Step-by-step explanation:

The relationship between tangent and cosine is ...

tan(α) = √(1/cos(α)² -1)

The cosine of the angle is given as 3/4, so the tangent is ...

tan(arccos(3/4)) = √(1/(3/4)² -1) = √(16/9 -1) = √(7/9)

tan(arccos(3/4)) = (√7)/3

6 0

3 years ago

5. While bird watching, Gabrielle saw 6 robins and 3 blue jays.

Semenov [28]

Answer:

Neither. They both saw the same amount of birds.

Step-by-step explanation:

Gabrielle saw 9 birds, but so did Chase. 6+3 is 9. 3+6 is also 9. The commutative property of addition states that a+b=b+a.

4 0

3 years ago

Given: AB || CD

jarptica [38.1K]

First, find slope of point A and B using slope formula:
y2-y1 = 7-0 = -7
x2-x1 3-8 5

Next, use the point-slope formula to find the equation (pick either point A or B to substitute into this equation; the answer will be the same either way):
y-y1=m(x-x1)
y-7=-7(x-3) (I used point B here)
5
y=-7x +56 so the y-intercept is 56. Hurray! Part 1 down!
 5 5 5

Now to answer part 2. Since AB ll CD, they have the same slope: -7
5
Therefore, you can use the handy point-slope equation to calculate the equation of line CD. (Remember you only need one of the points to use this equation if you already have the slope.) Since the only point given is D(5,5), we'll use that one:
y-y1=m(x-x1)
y-5=-7(x-5)
 5
y=-7x + 12 Yay! That's the answer to part 2; the equation of line CD
 5

3 0

3 years ago

The circular stream of water from a faucet is observed to taper from a diameter of 20 mm to 10 mm in a distance of 50 cm. Determ

steposvetlana [31]

Solution:

From Bernoulli equation

$\frac{1}{2}$ ρ $V_{1}^{2}$ + ρgh = $\frac{1}{2}$ ρ $V_{2}^{2}$ , where ρ is density of water, h – height difference and $V_{1}$ and $V_{2}$ are the velocities in upper and lower cross sections correspondingly. We take into account that the pressure in both cross-sections is the same and equal to the atmospheric one.

From the continuity and assuming water incompressible:

$\pi r_{1}^{2}$ $V_{1}$ = $\pi r_{2}^{2}$ $V_{2}$ , where $A_{1}$ and $A_{2}$ are the corresponding cross-sections. As the lower diameter is twice as low as the upper one, we can conclude that $V_{1} = \frac{V_{2}}{4}$

Inserting it back into Bernoulli equations produces:

$V_{1} = \sqrt{\frac{2}{15}gh} = 0.26m/s$ and the flow rate is

$\pi r_{1}^{2}V_{1} = \pi r_{2}^{2}V_{2} = 8.10^{-5} \frac{m^{3} }{s}$

6 0

3 years ago