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3 years ago

What percent of 88 is 154?

Mathematics

1 answer:

lesantik [10]3 years ago

5 0

Answer:

It Is 175% Of 88.

Step-by-step explanation:

154 ÷ 88 = 1.75

To Convert A Decimal Into Percent, You Multiply By 100.

1.75 · 100% = 175%

Therefore, The Answer Is 175%.

You might be interested in

What is the image point of (2,7) after the transformation r y=-x R270?

Semmy [17]

Answer:

Step-by-step explanation:|-2 - (-1)| =

|-2 + 1| =

| -1 | =

So, we're one unit away from y = -1. If we're one unit ABOVE y=-1, we need to move the point to be one unit BELOW it instead. And if we're one unit BELOW y = -1, we need to move the point to be one unit ABOVE it.

Since the point has a y-coordinate of -2, its BELOW the line y = -1, by 1 unit. Now we change the y-coordinate so its instead ABOVE y = -1, by 1 unit:

(y-coordinate of reflection line) + (number of units above that line) =

-1 + 1 =

Our new y-coordinate is 0, so the point is now at

(7, 0).

A more mechanical, but less intuitive approach is as follows:

Let L = the y-coordinate of the line, and

P = the y-coordinate of the point.

The new y-coordinate is 2L - P.

In this case, L = -1, and P = -2. So we have

2L - P =

2(-1) - (-2) =

-2 -(-2) =

5 0

2 years ago

What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.

lapo4ka [179]

Consider the equation $x^6+6x^3+5=0$ .

First, you can use the substitution $t=x^3$ , then $x^6=(x^3)^2=t^2$ and equation becomes $t^2+6t+5=0$ . This equation is quadratic, so

$D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1$ .

Then you can factor this equation:

$(t+5)(t+1)=0$ .

Use the made substitution again:

$(x^3+5)(x^3+1)=0$ .

You have in each brackets the expression like $a^3+b^3$ that is equal to $(a+b)(a^2-ab+b^2)$ . Thus,

$x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)$

and the equation is

$(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0$ .

Here $x_1=-\sqrt[3]{5} , x_2=-1$ and you can sheck whether quadratic trinomials have real roots:

1. $D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}$ .

2. $D_2=(-1)^2-4\cdot 1=1-4=-3$ .

This means that quadratic trinomials don't have real roots.

Answer: $x_1=-\sqrt[3]{5} , x_2=-1$

If you need complex roots, then

$x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2} ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}$ .

6 0

3 years ago

Determine whether the triangles are similar. or not.

mr Goodwill [35]

Answer:

Step-by-step explanation:

We know that ∠ $ABE$ ≅ ∠ $DBC$ because they are vertical angles, so that's one pair of congruent angles. However, we can't compare ∠ $BAE$ and ∠ $DCB$ as they have different measures, so we'll have to find the measure of either ∠ $AEB$ or ∠ $BDC$ . Let's go with the former. Using the fact that the sum of the measures of the interior angles of a triangle is $180$ °, we know that $m$ ∠ $AEB=180-(53+42)=180-95=85$ °. Now, we know that ∠ $AEB$ ≅ ∠ $DCB$ because their measures are equal. That's another pair of congruent angles. Therefore, the angles are similar by $AA$ . Hope this helps!