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3 years ago

7

There are 5 forks in the silverware drawer. There are 2 more knives than forks, and there are twice as many spoons as knives. Ho

w many pieces of silverware are there in all? PLEASE ANSWER QUICK

1 answer:

777dan777 [17]3 years ago

7 0

The answer to this question is 26

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Give an example of an absolute value inequality whose answers falls between two values. Give an example of an absolute value ine

mel-nik [20]

Answer:

A) Suppose we have an onedimensional situation.

in the 0 of our x-axis, we have a fruit tree, and we want to rest at a distance no bigger than 6 ft of the tree, then all the possible positions of our resting place are:

x ∈(-6ft, 6ft)

we can write this as: IxI < 6ft

b) now we think the opposite situation, we want to rest at least 6ft away from the tree, then we have that:

x ∉ [-6ft, 6ft].

or IxI > 6ft.

So you can see that the difference in those two cases is if we want to be "inside a given range" (for the first case) or "outside a given range" (for the second case).

8 0

3 years ago

Triangle ABC has vertices at A(−3, 4), B(4, −2), C(8, 3). The triangle translates 2 units up and 1 unit right. Which rule repres

Nataly [62]

Answer:

The translation statement is given by:

$(x,y)\rightarrow (x+1,y+2)$

After the translation, the coordinates of vertex A is (-2,6).

Step-by-step explanation:

Given :

Vertices of a triangle ABC are:

A(−3, 4), B(4, −2), C(8, 3)

The triangle is translated 2 units up and 1 unit right.

To find the co-ordinates of point A after translation.

Translation rules.

For shift of $c$ units up, the translation is given as:

$(x,y)\rightarrow (x,y+c)$

For shift of $k$ units right, the translation is given as:

$(x,y)\rightarrow (x+k,y)$

So, it says the triangles is translated 2 units up and 1 unit right.

The translation statement is given by:

$(x,y)\rightarrow (x+1,y+2)$

So, co-ordinates of point A after translation is given by :

$(-3,4)\rightarrow (-3+1,4+2)=(-2,6)$

After the translation, the coordinates of vertex A is (-2,6).

4 0

3 years ago

Write an equation in point slope form for the line that passes through (-5,6) and (-3,-9).

Finger [1]

$\bf (\stackrel{x_1}{-5}~,~\stackrel{y_1}{6})\qquad
(\stackrel{x_2}{-3}~,~\stackrel{y_2}{-9})
\\\\\\
slope = m\implies
\cfrac{\stackrel{rise}{ y_2- y_1}}{\stackrel{run}{ x_2- x_1}}\implies \cfrac{-9-6}{-3-(-5)}\implies \cfrac{-9-6}{-3+5}\implies -\cfrac{15}{2}$

$\bf \begin{array}{|c|ll}
\cline{1-1}
\textit{point-slope form}\\
\cline{1-1}
\\
y-y_1=m(x-x_1)
\\\\
\cline{1-1}
\end{array}\implies y-6=-\cfrac{15}{2}[x-(-5)]\implies y-6=-\cfrac{15}{2}(x+5)
\\\\\\
y-6=-\cfrac{15}{2}x-\cfrac{75}{2}\implies y=-\cfrac{15}{2}x-\cfrac{75}{2}-6\implies y=-\cfrac{15}{2}x-\cfrac{87}{2}$

8 0

3 years ago

Assignment: Wed : Finding Unit Rates (7.RP.1, 7. RP.2, 7.RP.2a

spayn [35]

Answer:

5

Step-by-step explanation:

6 0

3 years ago

A ball is thrown into the air by a baby alien on a planet in the system of Alpha Centauri with a velocity of 30 ft/s. Its height

Crank

Answer:

a) h = 0.1: $\bar v = -11\,\frac{ft}{s}$ , h = 0.01: $\bar v = -10.1\,\frac{ft}{s}$ , h = 0.001: $\bar v = -10\,\frac{ft}{s}$ , b) The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

Step-by-step explanation:

a) We know that $y = 30\cdot t -10\cdot t^{2}$ describes the position of the ball, measured in feet, in time, measured in seconds, and the average velocity ( $\bar v$ ), measured in feet per second, can be done by means of the following definition:

$\bar v = \frac{y(2+h)-y(2)}{h}$

Where:

$y(2)$ - Position of the ball evaluated at $t = 2\,s$ , measured in feet.

$y(2+h)$ - Position of the ball evaluated at $t =(2+h)\,s$ , measured in feet.

$h$ - Change interval, measured in seconds.

Now, we obtained different average velocities by means of different change intervals:

$h = 0.1\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.1) = 30\cdot (2.1)-10\cdot (2.1)^{2}$

$y(2.1) = 18.9\,ft$

$\bar v = \frac{18.9\,ft-20\,ft}{0.1\,s}$

$\bar v = -11\,\frac{ft}{s}$

$h = 0.01\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.01) = 30\cdot (2.01)-10\cdot (2.01)^{2}$

$y(2.01) = 19.899\,ft$

$\bar v = \frac{19.899\,ft-20\,ft}{0.01\,s}$

$\bar v = -10.1\,\frac{ft}{s}$

$h = 0.001\,s$

$y(2) = 30\cdot (2) - 10\cdot (2)^{2}$

$y (2) = 20\,ft$

$y(2.001) = 30\cdot (2.001)-10\cdot (2.001)^{2}$

$y(2.001) = 19.99\,ft$

$\bar v = \frac{19.99\,ft-20\,ft}{0.001\,s}$

$\bar v = -10\,\frac{ft}{s}$

b) The instantaneous velocity when $t = 2\,s$ can be obtained by using the following limit:

$v(t) = \lim_{h \to 0} \frac{x(t+h)-x(t)}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot (t+h)-10\cdot (t+h)^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h -10\cdot (t^{2}+2\cdot t\cdot h +h^{2})-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot t +30\cdot h-10\cdot t^{2}-20\cdot t \cdot h-10\cdot h^{2}-30\cdot t +10\cdot t^{2}}{h}$

$v(t) = \lim_{h \to 0} \frac{30\cdot h-20\cdot t\cdot h-10\cdot h^{2}}{h}$

$v(t) = \lim_{h \to 0} 30-20\cdot t-10\cdot h$

$v(t) = 30\cdot \lim_{h \to 0} 1 - 20\cdot t \cdot \lim_{h \to 0} 1 - 10\cdot \lim_{h \to 0} h$

$v(t) = 30-20\cdot t$

And we finally evaluate the instantaneous velocity at $t = 2\,s$ :

$v(2) = 30-20\cdot (2)$

$v(2) = -10\,\frac{ft}{s}$

The instantaneous velocity of the ball when $t = 2\,s$ is $-10$ feet per second.

8 0

3 years ago