zolas mother spends 39 minutes braiding zolas hair she finishes at 7:52 what time did she start braiding zolas hair

Basile [38]

Answer:

$x=8\frac34\ \,,\quad y=8\frac12$

Step-by-step explanation:

$\Delta TRS\sim \Delta LJK \quad\implies\quad \dfrac{|TR|}{|LJ|}=\dfrac{|RS|}{|JK|}=\dfrac{|ST|}{|KL|}\\\\\dfrac{4}{7}=\dfrac{5}{x}=\dfrac{6}{y+2}\\\\\\\dfrac{4}{7}=\dfrac{5}{x}\quad\implies\quad4x=5\cdot7\quad\implies\quad x=\dfrac{35}4=8\frac34\\\\\\\dfrac{5}{x}=\dfrac{6}{y+2}\quad\implies\quad 5(y+2)=6\cdot\dfrac{35}4\\\\5y+10=3\cdot\dfrac{35}2\\\\5y\ =\ \dfrac{3\cdot35}2-10\\ \div5\qquad\quad \div5\\\\y=\dfrac{3\cdot7}2-2=10\frac12-2=8\frac12$

3 0

3 years ago

Rounding factors to the greatest place results in this

san4es73 [151]

Estimate is the answer

3 0

2 years ago

Problem 10: A tank initially contains a solution of 10 pounds of salt in 60 gallons of water. Water with 1/2 pound of salt per g

AysviL [449]

Answer:

The quantity of salt at time t is $m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$ , where t is measured in minutes.

Step-by-step explanation:

The law of mass conservation for control volume indicates that:

$\dot m_{in} - \dot m_{out} = \left(\frac{dm}{dt} \right)_{CV}$

Where mass flow is the product of salt concentration and water volume flow.

The model of the tank according to the statement is:

$(0.5\,\frac{pd}{gal} )\cdot \left(6\,\frac{gal}{min} \right) - c\cdot \left(6\,\frac{gal}{min} \right) = V\cdot \frac{dc}{dt}$

Where:

$c$ - The salt concentration in the tank, as well at the exit of the tank, measured in $\frac{pd}{gal}$ .

$\frac{dc}{dt}$ - Concentration rate of change in the tank, measured in $\frac{pd}{min}$ .

$V$ - Volume of the tank, measured in gallons.

The following first-order linear non-homogeneous differential equation is found:

$V \cdot \frac{dc}{dt} + 6\cdot c = 3$

$60\cdot \frac{dc}{dt} + 6\cdot c = 3$

$\frac{dc}{dt} + \frac{1}{10}\cdot c = 3$

This equation is solved as follows:

$e^{\frac{t}{10} }\cdot \left(\frac{dc}{dt} +\frac{1}{10} \cdot c \right) = 3 \cdot e^{\frac{t}{10} }$

$\frac{d}{dt}\left(e^{\frac{t}{10}}\cdot c\right) = 3\cdot e^{\frac{t}{10} }$

$e^{\frac{t}{10} }\cdot c = 3 \cdot \int {e^{\frac{t}{10} }} \, dt$

$e^{\frac{t}{10} }\cdot c = 30\cdot e^{\frac{t}{10} } + C$

$c = 30 + C\cdot e^{-\frac{t}{10} }$

The initial concentration in the tank is:

$c_{o} = \frac{10\,pd}{60\,gal}$

$c_{o} = 0.167\,\frac{pd}{gal}$

Now, the integration constant is:

$0.167 = 30 + C$

$C = -29.833$

The solution of the differential equation is:

$c(t) = 30 - 29.833\cdot e^{-\frac{t}{10} }$

Now, the quantity of salt at time t is:

$m_{salt} = V_{tank}\cdot c(t)$

$m_{salt} = (60)\cdot (30 - 29.833\cdot e^{-\frac{t}{10} })$

Where t is measured in minutes.

7 0

3 years ago

Which of the following is true about a parallelogram? A. Opposite angles of a parallelogram are not congruent. B. Parallelograms

mafiozo [28]

Answer: The option is D.

Step-by-step explanation:

A line that intersects another line segment and separates it into two equal parts is called a bisector.

In a quadrangle, the line connecting two opposite corners is called a diagonal. We will show that in a parallelogram, each diagonal bisects the other diagonal.

Problem

ABCD is a parallelogram, and AC and BD are its two diagonals. Show that AO = OC and that BO = OD

Strategy

Once again, since we are trying to show line segments are equal, we will use congruent triangles. And here, the triangles practically present themselves. Let’s start with showing that AO is equal in length to OC, by using the two triangles in which AO and OC are sides: ΔAOD and ΔCOB.

There are all sorts of equal angles here that we can use. Several pairs of (equal) vertical angles, and several pairs of alternating angles created by a transversal line intersecting two parallel lines. So finding equal angles is not a problem. But we need at least one side, in addition to the angles, to show congruency.

As we have already proven, the opposite sides of a parallelogram are equal in size, giving us our needed side.

Once we show that ΔAOD and ΔCOB are congruent, we will have the proof needed, not just for AO=OC, but for both diagonals, since BO and OD are alsocorresponding sides of these same congruent triangles.

ABCD is a parallelogram

Given

AD || BC

From the definition of a parallelogram

AD = BC Opposite sides of a parallelogram are equal in size

∠OBC ≅ ∠ODA Alternate Interior Angles Theorem, ∠OCB ≅ ∠OAD Alternate Interior Angles Theorem,

ΔOBC ≅ ΔODA

Angle-Side-Angle

BO=OD Corresponding sides in congruent triangles AO=OC Corresponding sides in congruent triangles.

5 0

3 years ago

What is happening to this graph when the x values are between -1 and 1

Amiraneli [1.4K]

It is decreasing, since the y-value is going down.

3 0

3 years ago

Read 2 more answers