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patriot [66]
3 years ago
15

1 7/8 divided by 3/4 can someone please help me

Mathematics
2 answers:
antoniya [11.8K]3 years ago
5 0

Answer:

1 7/8

Step-by-step explanation:

sesenic [268]3 years ago
3 0
1. Change 1 7/8 to an improper fraction, which gives you 15/8.

2. Use KFC to divide fractions. KEEP the first term, FLIP the second and CHANGE to multiplication.

3. 15/8 x 4/3 = 60/24 = 2 1/2
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Please I need the answer PLEASEEEEEEEEEEEEEEEEEE!
Ilya [14]

Answer: The first one

Step-by-step explanation:

3 0
3 years ago
Strain-displacement relationship) Consider a unit cube of a solid occupying the region 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, 0 ≤ z ≤ 1 After loa
Anastasy [175]

Answer:

please see answers are as in the explanation.

Step-by-step explanation:

As from the data of complete question,

0\leq x\leq 1\\0\leq y\leq 1\\0\leq z\leq 1\\u= \alpha x\\v=\beta y\\w=0

The question also has 3 parts given as

<em>Part a: Sketch the deformed shape for α=0.03, β=-0.01 .</em>

Solution

As w is 0 so the deflection is only in the x and y plane and thus can be sketched in xy plane.

the new points are calculated as follows

Point A(x=0,y=0)

Point A'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point A'(0+<em>(0.03)</em><em>(0),0+</em><em>(-0.01)</em><em>(0))</em>

Point A'(0<em>,0)</em>

Point B(x=1,y=0)

Point B'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point B'(1+<em>(0.03)</em><em>(1),0+</em><em>(-0.01)</em><em>(0))</em>

Point <em>B</em>'(1.03<em>,0)</em>

Point C(x=1,y=1)

Point C'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point C'(1+<em>(0.03)</em><em>(1),1+</em><em>(-0.01)</em><em>(1))</em>

Point <em>C</em>'(1.03<em>,0.99)</em>

Point D(x=0,y=1)

Point D'(x+<em>α</em><em>x,y+</em><em>β</em><em>y) </em>

Point D'(0+<em>(0.03)</em><em>(0),1+</em><em>(-0.01)</em><em>(1))</em>

Point <em>D</em>'(0<em>,0.99)</em>

So the new points are A'(0,0), B'(1.03,0), C'(1.03,0.99) and D'(0,0.99)

The plot is attached with the solution.

<em>Part b: Calculate the six strain components.</em>

Solution

Normal Strain Components

                             \epsilon_{xx}=\frac{\partial u}{\partial x}=\frac{\partial (\alpha x)}{\partial x}=\alpha =0.03\\\epsilon_{yy}=\frac{\partial v}{\partial y}=\frac{\partial ( \beta y)}{\partial y}=\beta =-0.01\\\epsilon_{zz}=\frac{\partial w}{\partial z}=\frac{\partial (0)}{\partial z}=0\\

Shear Strain Components

                             \gamma_{xy}=\gamma_{yx}=\frac{\partial u}{\partial y}+\frac{\partial v}{\partial x}=0\\\gamma_{xz}=\gamma_{zx}=\frac{\partial u}{\partial z}+\frac{\partial w}{\partial x}=0\\\gamma_{yz}=\gamma_{zy}=\frac{\partial w}{\partial y}+\frac{\partial v}{\partial z}=0

Part c: <em>Find the volume change</em>

<em></em>\Delta V=(1.03 \times 0.99 \times 1)-(1 \times 1 \times 1)\\\Delta V=(1.0197)-(1)\\\Delta V=0.0197\\<em></em>

<em>Also the change in volume is 0.0197</em>

For the unit cube, the change in terms of strains is given as

             \Delta V={V_0}[(1+\epsilon_{xx})]\times[(1+\epsilon_{yy})]\times [(1+\epsilon_{zz})]-[1 \times 1 \times 1]\\\Delta V={V_0}[1+\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}+\epsilon_{xx}\epsilon_{zz}+\epsilon_{yy}\epsilon_{zz}+\epsilon_{xx}\epsilon_{yy}\epsilon_{zz}-1]\\\Delta V={V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\

As the strain values are small second and higher order values are ignored so

                                      \Delta V\approx {V_0}[\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\ \Delta V\approx [\epsilon_{xx}+\epsilon_{yy}+\epsilon_{zz}]\\

As the initial volume of cube is unitary so this result can be proved.

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3 years ago
Use the definition mtan=lim
Leokris [45]

(a) mtan refers to the slope of the tangent line. Given <em>f(x)</em> = 9 + 7<em>x</em> ², compute the difference quotient:

\dfrac{f(x+h)-f(x)}h = \dfrac{(9+7(x+h)^2)-(9+7x^2)}h = \dfrac{(9+7x^2+14xh+7h^2)-9-7x^2}h = \dfrac{14xh+7h^2}h

Then as <em>h</em> approaches 0 - bearing in mind that we're specifically considering <em>h</em> <em>near</em> 0, and not <em>h</em> = 0 - we can eliminate the factor of <em>h</em> in the numerator and denominator, so that

m_{\rm tan} = \displaystyle \lim_{h\to0}\frac{f(x+h)-f(x)}h = \lim_{h\to0}\frac{14xh+7h^2}h = \lim_{h\to0}(14x+7h) = 14x

and so the slope of the line at <em>P</em> (0, 9), for which we take <em>x</em> = 0, is 0.

(b) The equation of the tangent line is then <em>y</em> = 9.

4 0
2 years ago
TOP OF THE MORNIN TO THE PEOPLE THAT CAN ANSWER THIS!!!!
Dvinal [7]
<span>surface area=4π<span>r2</span>=2700 c<span>m2</span></span>
<span><span>r2</span>=<span>2700<span>4π</span></span>=<span>2700<span>4×3</span></span>=225</span>
<span>r=<span>225<span>−−−</span>√</span>=15</span>
<span><span>volume=<span>43</span>π<span>r3</span>=<span>43</span>×3×<span>153</span>=13500 c<span>m3</span></span></span>
7 0
3 years ago
After 4 days, a bookstore has 140 copies of a new title still on hand. After 9 days, the bookstore has 50 copies still on hand.
alexira [117]

Answer:

c(t) =-18t +212

Step-by-step explanation:

Given

(t_1,c_1) = (4,140)

(t_2,c_2) = (9,50)

Required

Determine the equation

First, we calculate the slope (m)

m = \frac{c_2 - c_1}{t_2 - t_1}

m = \frac{50 - 140}{9-4}

m = \frac{-90}{5}

m = -18

The equation is then calculated as:

c =m(t - t_1) + c_1

c =-18*(t - 4) + 140

c =-18t +72 + 140

c =-18t +212

Hence, the function is:

c(t) =-18t +212

8 0
3 years ago
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