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3 years ago

HELP!!!!

2 answers:

8 0

A: (5(6))/(3(-4)+(6)) = -5
B: 2(6)^2-(-4)^3 = 136

Substitute the value given as x and y into your expression. For A. , order of operation matters. I assumed the “+x” is part of the denominator.

Law Incorporation [45]3 years ago

5 0

Answer:

A. 5x/3y+x=5×6/(3×-4)=30/-12=-10/4=-5/2

B. 2x (square) - y (cube)=2×6²-(-4)³=72--64=72+64=136

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HELP ASP ON KITCHEN MATH !!!

Komok [63]

Answer:

1 1/2 shortening

1 tsp salt

4 1/2 granulated sugar

1 or 3 tsp baking soda (not sure if it’s 1/2 or 1 1/2 for the original recipe)

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8 0

3 years ago

Read 2 more answers

What is the lateral area of the cone? 10 yd and 12.1 yd

Harman [31]

Answer:

The lateral area will be "381 yd²". A further solution is provided below.

Step-by-step explanation:

The given values are:

Radius of cone,

r = 10 yd

Height of cone,

L = 12.1 yd

As we know,

The lateral surface area of cone is:

= $\pi r L$

On substituting the values, we get

= $\frac{22}{7}\times 10\times 12.1$

= $\frac{22}{7}\times 121$

= $380.28 \ yd^2$

0r,

= $381 \ yd^2$

8 0

3 years ago

Write (8a-^3) -2/3 in simplest form

Anna71 [15]

$(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

Solution:

Given that,

$(8a^{-3})^{\frac{-2}{3}$

We have to write in simplest form

Use the following law of exponent

$(a^m)^n = a^{mn}$

Using this, simplify the given expression

$(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^{ -3 \times \frac{-2}{3}}\\\\Simplifying\ we\ get\\\\(8a^{-3})^{\frac{-2}{3}} = 8^{\frac{-2}{3}} \times a^2\\\\We\ know\ that\ 8 = 2^3\\\\Therefore\\\\(8a^{-3})^{\frac{-2}{3}} =2^3^{\frac{-2}{3}} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} =2^{-2} \times a^2\\\\(8a^{-3})^{\frac{-2}{3}} = \frac{a^2}{4}$

Thus the given expression is simplified

8 0

3 years ago

Write a quadratic equation with the given roots. Write the equation in the form of ax^2+bx+c=0 where a b and c are integers

Bas_tet [7]

You haven't provided the required roots, but I can tell you how to do this kind of exercises in general.

If the $x^2$ coefficient is 1, i.e. the equation is written like $x^2+bx+c=0$ , then you can say the following about the coefficients b and c:

$b$ is the opposite of the sum of the roots
$c$ is the multiplication of the roots.

So, for example, if we want an equation whose roots are 4 and -2, we have:

$4+(-2) = 4-2 = 2 \implies b = -2$
$4 \cdot (-2) = -8 \implies c = -8$

So, the equation is $x^2-2x-8=0$

If your roots are rational, you can work like this: suppose you want an equation with roots 3/4 and 1/2. You have:

$\dfrac{3}{4}+\dfrac{1}{2} = \dfrac{3}{4}+\dfrac{2}{4} = \dfrac{5}{4} \implies b = -\dfrac{5}{4}$
$\dfrac{3}{4} \cdot \dfrac{1}{2} = \dfrac{3}{8} \implies c = \dfrac{3}{8}$