Emilio bakes 2 pies. he shares them equally among 3 friends. how many pies does each person get?

Which figure has reflection symmetry?

lyudmila [28]

Answer:

the arrow :)

Step-by-step explanation:

3 0

4 years ago

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What is 6x to the 3 fifths in radical form

DerKrebs [107]

Assuming you mean $6x^ \frac{3}{5}$ and not $(6x)^ \frac{3}{5}$

remember
$x^ \frac{m}{n}= \sqrt[n]{x^m}$ so

$6x^ \frac{3}{5}$ = $6 \sqrt[5]{x^3}$

6 0

4 years ago

Find the product (3a4+4)2

asambeis [7]

The result is

9

a

2

−

16

The reason is the following:

The problem is an example of a notable product: "the sum multiplied by the diference is equal to the difference of squares", that is to say:

(

a

+

b

)

⋅

(

a

−

b

)

=

a

2

−

b

2

.

By applying this to our question, we obtain that:

(

3

a

−

4

)

⋅

(

3

a

+

4

)

=

(

3

a

)

2

−

(

4

)

2

=

9

a

2

−

16

.

4 0

4 years ago

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Looking at the top of tower A and base of tower B from points C and D, we find that ∠ACD = 60°, ∠ADC = 75° and ∠ADB = 30°. Let t

katrin2010 [14]

Answer:

$\text{Exact: }AB=25\sqrt{6},\\\text{Rounded: }AB\approx 61.24$

Step-by-step explanation:

We can use the Law of Sines to find segment AD, which happens to be a leg of $\triangle ACD$ and the hypotenuse of $\triangle ADB$ .

The Law of Sines states that the ratio of any angle of a triangle and its opposite side is maintained through the triangle:

$\frac{a}{\sin \alpha}=\frac{b}{\sin \beta}=\frac{c}{\sin \gamma}$

Since we're given the length of CD, we want to find the measure of the angle opposite to CD, which is $\angle CAD$ . The sum of the interior angles in a triangle is equal to 180 degrees. Thus, we have:

$\angle CAD+\angle ACD+\angle CDA=180^{\circ},\\\angle CAD+60^{\circ}+75^{\circ}=180^{\circ},\\\angle CAD=180^{\circ}-75^{\circ}-60^{\circ},\\\angle CAD=45^{\circ}$

Now use this value in the Law of Sines to find AD:

$\frac{AD}{\sin 60^{\circ}}=\frac{100}{\sin 45^{\circ}},\\\\AD=\sin 60^{\circ}\cdot \frac{100}{\sin 45^{\circ}}$

Recall that $\sin 45^{\circ}=\frac{\sqrt{2}}{2}$ and $\sin 60^{\circ}=\frac{\sqrt{3}}{2}$ :

$AD=\frac{\frac{\sqrt{3}}{2}\cdot 100}{\frac{\sqrt{2}}{2}},\\\\AD=\frac{50\sqrt{3}}{\frac{\sqrt{2}}{2}},\\\\AD=50\sqrt{3}\cdot \frac{2}{\sqrt{2}},\\\\AD=\frac{100\sqrt{3}}{\sqrt{2}}\cdot\frac{ \sqrt{2}}{\sqrt{2}}=\frac{100\sqrt{6}}{2}={50\sqrt{6}}$

Now that we have the length of AD, we can find the length of AB. The right triangle $\triangle ADB$ is a 30-60-90 triangle. In all 30-60-90 triangles, the side lengths are in the ratio $x:x\sqrt{3}:2x$ , where $x$ is the side opposite to the 30 degree angle and $2x$ is the length of the hypotenuse.

Since AD is the hypotenuse, it must represent $2x$ in this ratio and since AB is the side opposite to the 30 degree angle, it must represent $x$ in this ratio (Derive from basic trig for a right triangle and $\sin 30^{\circ}=\frac{1}{2}$ ).

Therefore, AB must be exactly half of AD:

$AB=\frac{1}{2}AD,\\AB=\frac{1}{2}\cdot 50\sqrt{6},\\AB=\frac{50\sqrt{6}}{2}=\boxed{25\sqrt{6}}\approx 61.24$

3 0

3 years ago

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Tres parcelas de tierras de cultivo están a la venta. Cada parcela es de diferente tamaño. Parcela Area (en acres) Precio (en dó

Serhud [2]

Answer:

yes ok

Step-by-step explanation:

8 0

3 years ago