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2 years ago

What is forty-five divided by seven? WILL MARK BRAINLIEST

Mathematics

2 answers:

nalin [4]2 years ago

6 0

Answer:

6.4285714286

Step-by-step explanation:

Evgesh-ka [11]2 years ago

6 0

Answer:

6 and 3/7 (6.428571428..............) Almost 6.5

Step-by-step explanation:

45/7 is an improper fraction

Convert in mixed number

6(3/7)

When you multiply 7*6 = 42.But from 42 to 45 there is 3. You put 3 on the top,and 7 at the bottom.So you have 6 whole sevens and 3/7 (almost a half of 7)

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michael basketball team practice fir 2 hours and 40 minutes yesturday and 3 hours 15 minutes today.how much longer did team prac

Serjik [45]

Answer:

35 minutes

Step-by-step explanation:

2 hrs 4 min= 160 min

3 hrs 15 min= 195 min

195-160=35

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3 years ago

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svetlana [45]

Answer:

3 $\sqrt{2}$

Step-by-step explanation:

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3 0

3 years ago

I’m confused on this one

erastovalidia [21]

In the chart of the triangle it shows that the variable b is the base. It also give us the equation for area 1/2bh (half of base * height)

Now we just plug in the values we are given

1/2bh

1/2(10*13)

1/2(130)

The area of the triangle is 65 ft

4 0

2 years ago

Read 2 more answers

At an archeological site, the remains of two ancient step pyramids are congruent. If ABCD is congruent to EFGH, find GH

Alona [7]

Answer:

40 ft

Step-by-step explanation:

If two polygons are congruent to one another, therefore, it implies that the angles and sides of one is congruent to the corresponding sides and angles of the other.

Given that ABCD is congruent to EFGH, therefore:

CD ≅ GH

Since CD = 40 ft, therefore, GH will be 40 ft.

8 0

2 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.