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4 years ago

13

which is the greatest out of 1\3, -1.2, 0.45, & -4/5 record your answer using correct place value in the griddable *show wor

k please*

1 answer:

Andrej [43]4 years ago

5 0

The answer is 1/2 because if you draw 1/2 and 1/3 I will see 1/2 is bigger

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F (x) = - eˣ Baseline (x - 4)<br> What is(are) the critical point(s) of f?

Ber [7]

Answer

given,

$f(x) = \dfrac{-e^x}{x - 4}$

to find the critical point of the given expression

fist differentiating the function

$f'(x) = -\dfrac{(x-4)e^x+ e^x}{(x - 4)^2}$

$f'(x) = \dfrac{-(x-4)e^x- e^x}{(x - 4)^2}$

$f'(x) = \dfrac{-e^x(x-3)}{(x - 4)^2}$

now equating differential equation to zero

$\dfrac{e^x(-x+3)}{(x - 4)^2}=0$

$e^x(-x+3)=0$

now,

-x + 3 = 0 and eˣ ≠ 0

x = 3

the critical number will be equal to x = 3

$y = \dfrac{-e^3}{3 - 4}$

$y =e^3$

8 0

4 years ago

5/6 plus 3/4 plus 2/3 <br> Equals?

Ann [662]

Answer:

9/4

Step-by-step explanation:

4 0

3 years ago

Subtract -1 3/5 - ( -2 7/8) Enter your answer as a simplified fraction by filling in the boxes

vredina [299]

The answer is 51/40=1 and 11/40

8 0

3 years ago

A piece of wire 19 m long is cut into two pieces. One piece is bent into a square and the other is bent into an equilateral tria

mr Goodwill [35]

Answer: 8.26 m

Step-by-step explanation:

$Let s be the length of the wire used for the square. \\Let $t$ be the length of the wire used for the triangle. \\Let $A_{S}$ be the area of the square. \\Let ${A}_{T}}$ be the area of the triangle. \\One side of the square is $\frac{s}{4}$ \\Therefore,we know that,$$A_{S}=\left(\frac{s}{4}\right)^{2}=\frac{s^{2}}{16}$

$The formula for the area of an equilateral triangle is, $A=\frac{\sqrt{3}}{4} a^{2}$ where $a$ is the length of one side,And one side of our triangle is $\frac{t}{3}$So,We know that,$$A_{T}=\frac{\sqrt{3}}{4}\left(\frac{t}{3}\right)^{2}$$We have to find the value of "s" such that,$\mathrm{s}+\mathrm{t}=19$ hence, $\mathrm{t}=19-\mathrm{s}$And$$A_{S}+A_{T}=A_{S+T}$

$$$Therefore,$$\begin{aligned}&A_{T}=\frac{\sqrt{3}}{4}\left(\frac{(19-s)}{3}\right)^{2}=\frac{\sqrt{3}(19-s)^{2}}{36} \\&A_{T+S}=\frac{s^{2}}{16}+\frac{\sqrt{3}(19-s)^{2}}{36}\end{aligned}$

$Differentiating the above equation with respect to s we get,$$A^{\prime}{ }_{T+S}=\frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}$$Now we solve $A_{S+T}^{\prime}=0$$$\begin{aligned}&\Rightarrow \frac{s}{8}-\frac{\sqrt{3}(19-s)}{18}=0 \\&\Rightarrow \frac{s}{8}=\frac{\sqrt{3}(19-s)}{18}\end{aligned}$$Cross multiply,$$\begin{aligned}&18 s=8 \sqrt{3}(19-s) \\&18 s=152 \sqrt{3}-8 \sqrt{3} s \\&(18+8 \sqrt{3}) s=152 \sqrt{3} \\&s=\frac{152 \sqrt{3}}{(18+8 \sqrt{3})} \approx 8.26\end{aligned}$

$The domain of $s$ is $[0,19]$.So the endpoints are 0 and 19$$\begin{aligned}&A_{T+S}(0)=\frac{0^{2}}{16}+\frac{\sqrt{3}(19-0)^{2}}{36} \approx 17.36 \\&A_{T+S}(8.26)=\frac{8.26^{2}}{16}+\frac{\sqrt{3}(19-8.26)^{2}}{36} \approx 9.81 \\&A_{T+S}(19)=\frac{19^{2}}{16}+\frac{\sqrt{3}(19-19)^{2}}{36}=22.56\end{aligned}$

$$$Therefore, for the minimum area, $8.26 \mathrm{~m}$ should be used for the square$

8 0

2 years ago

Mario is making a diagram that shows the relationship between diferent kinds of quadrilaterals. in the diagram each quadrilatera

marissa [1.9K]

Yes it can becuase it is a quadrilateral

7 0

3 years ago

Read 2 more answers