PLS HELP ASAP IM TAKING AN RXAM!!! 40 POINTS!!!!!!
Using data from the Internet, we've compiled a list of Presidents of the United States through 2012 to show party affiliations and ages during office. You completed a frequency table and bar graph using this data. The bar graph is shown below. Describe what you can say about this bar graph given the data and the graph.

Answer:

1. 0.74 = 74%

4. 0.617 = 61.7

6. 68% = 0.68

9. 175% = 1.75

2. 0.15 = 15%

5. 0.834 = 83.4%

7. 34% = 0.34

10. 200% = 2.00

3. 0.09 = 9%

8. 58.5% = 0.585

Step-by-step explanation:

What are you asking?

Answer:

5/4 and 7/4 are the two solutions.

Answer:

E

Step-by-step explanation:

Like and give 5 star rating

Answer:

93.84

Step-by-step explanation:

I think I'm not sure

Step-by-step explanation:

use SOHCAHTOA as shown above in the picture

Answer:

33

Step-by-step explanation:

Answer:

62

Step-by-step explanation:

SA=2(5x2)+2(5x3)+2(3x2)

SA=20+30+12

SA=62

Answer:

y=2x-2

You use rise/run in this problem

Answer:

x = 6

Step-by-step explanation:

Because we have parallel lines, we can use alternate angles are equal. Therefore, we know that 11x-6 = 10x.

Minus 11 from both sides, and we get -6 = -x, so x = 6

[tex]\begin{array}{ccll} Juice&OJ\\ \cline{1-2} \frac{2}{3}&\frac{1}{3}\\[1em] x&1 \end{array}\implies \cfrac{~~ \frac{2}{3}~~}{x}=\cfrac{~~\frac{1}{3} ~~}{1}\implies \cfrac{~~ \frac{2}{3}~~}{x}=\cfrac{1}{3}\implies \cfrac{~~ \frac{2}{3}~~}{\frac{x}{1}}=\cfrac{1}{3} \\\\\\ \cfrac{2}{3x}=\cfrac{1}{3}\implies 6=3x\implies \cfrac{6}{3}=x\implies 2=x[/tex]

Answer

1. ΔABC ≅ ΔEFG

2. ΔMKL ≅ ΔVKU

3. ΔMLN ≅ ΔNCD (With this one I couldn't fully read the letter in the middle. I couldn't tell if it was an N or a B so just swap it for the right letter. <3)

4. ΔFGH ≅ ΔRHQ

5. ΔUVL ≅ ΔMNL

6. ΔQRS ≅ ΔVUT

Step-by-step explanation:

hope this helps. . .<3

good luck! UwU

Answer:

where is the problem to solve. please tell me

Answer:

y=1/3x-7/2

Step-by-step explanation:

First get the 2y by itself by adding 2/3x to both sides. Then subtract both sides by 2.

Answer:

I think it'

is A and B

Step-by-step explanation:

hope hwlps

Answer:

The traditional scale consists of two plates or bowls suspended at equal distances from a fulcrum. One plate holds an object of unknown mass (or weight), while known masses are added to the other plate until static equilibrium is achieved and the plates level off, which happens when the masses on the two plates are equal. The perfect scale rests at neutral. A

Answer:

yes, the triangle are congruent by SAS

Answer:

  x = -2

Step-by-step explanation:

The expression f(x) = 1 corresponds to the point on the graph that is 1 unit above the x-axis. You can locate that horizontal line and see where it intersects the graph. That point of intersection  is 2 grid squares to the left of the y-axis, where x = -2.

  f(-2) = 1

  x = -2

Answer:

Your answer would be x=2±√5

Step-by-step explanation:

Answer:

Step-by-step explanation:

First get the (m) which is the gradient by using the formula y2-y1/x2-x1

Pick any random 2 values from the table. I will pick. (1,4) (2,8)

Y2=8  Y1=4   X2=2  X1=1

8-4/2-1    4/1  = 4

Y=4x+c

Answer:

240oz.

Step-by-step explanation:

15oz=1batch

She needs 16 batches.

15oz. x 16batches = 240oz.

Hope this helps :)

Let

[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots[/tex]

Differentiating twice gives

[tex]\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots[/tex]

[tex]\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n[/tex]

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

[tex]\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0[/tex]

[tex]\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0[/tex]

Then the coefficients in the power series solution are governed by the recurrence relation,

[tex]\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}[/tex]

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

[tex]k=0 \implies n=0 \implies a_0 = a_0[/tex]

[tex]k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}[/tex]

[tex]k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}[/tex]

[tex]k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}[/tex]

It should be easy enough to see that

[tex]a_{n=2k} = \dfrac{a_0}{(2k)!}[/tex]

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

[tex]k = 0 \implies n=1 \implies a_1 = a_1[/tex]

[tex]k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}[/tex]

[tex]k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}[/tex]

[tex]k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}[/tex]

so that

[tex]a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}[/tex]

So, the overall series solution is

[tex]\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)[/tex]

[tex]\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}[/tex]

Answer:

Step-by-step explanation:

Answer:

y = -2x + 1

Step-by-step explanation:

Substitute  the slope and the given point into y = mx + b and solve for "b".

y = mx + b

7 = -2(-3) + b

7 = 6 + b

1 = b       the y-intercept is 1

The equation of the line is y = -2x + 1

Well, 30+23+23 = 76 which means there are only a total of 76 bags that are accounted for. 110-76 = 34 bags which we do not know if they have zips or not. If we assume they have zips, that would be 34+23 bags with zips which is 57 bags.

Hello!

-5(x+2)=35

-5x-10=35

-5x=35+10

-5x=45

x=-9

Hope it helps!

~Just a determined gal

#CarryOnGainingKnowledge

[tex]MysteriousNature[/tex]

Answer:

x = -9

Step-by-step explanation:

-5(x + 2) = 35

~Distribute

(-5 * x) + (-5 * 2) = 35

~Simplify

-5x - 10 = 35

~Add 10 to both sides

-5x = 45

~Divide -5 to both sides

x = -9

Best of Luck!

Usando técnicas de conteo, se encuentra que

1. Un cliente puede ordenar una comida de 784 formas.

2. 420 grupos de 4 letras se pueden formar con las letras de la palabra TECNICA.

3. 840 números de 4 dígitos se pueden formar con los primeros 7 números naturales.

4. 10 partidos distintos se pueden realizar.

5. Pueden colocarse de 3,628,800 formas.

6. 1260 señales distintas pueden indicarse.

Item 1:

La técnica usada es el principio fundamental de conteo, que afirma que si hay n cosas, cada una con [tex]n_1, n_2, \cdots, n_n[/tex] maneras de seren realizadas, el número total de maneras de ser realizadas es:

[tex]N = n_1 \times n_2 \times \cdots \times n_n[/tex]

En este problema, [tex]n_1 = 14, n_2 = 8, n_3 = 7[/tex], por eso:

[tex]T = 14 \times 8 \times 7 = 784[/tex]

Un cliente puede ordenar una comida de 784 formas.

Item 2:

El orden es importante, ya que TECN es una palabra diferente de NCET, por lo tanto, la fórmula de permutaciones se usa para resolver este problema.

Fórmula de permutaciociones:

El número de permutaciones de x elementos en un conjunto de n elementos es dada por:

[tex]P_{n,x} = \frac{n!}{(n - x)!}[/tex]

En este problema, 4 letras de 7, enconteces:

[tex]P_{7,4} = \frac{7!}{3!} = 840[/tex]

La letra C se repite dos veces, o sea:

[tex]T = \frac{P_{7,4}}{2} = \frac{840}{2} = 420[/tex]

420 grupos de 4 letras se pueden formar con las letras de la palabra TECNICA.

Item 3:

Permutaciones de 4 dígitos de 7, sin repeticiones, o sea:

[tex]P_{7,4} = \frac{7!}{3!} = 840[/tex]

840 números de 4 dígitos se pueden formar con los primeros 7 números naturales.

Item 4:

El orden no es importante, ya que Time 1 x Time 2 es la misma partida de Time 2 x Time 1, por lo tanto, la fórmula de combinaciones se usa para resolver este problema.

Fórmula de combinaciones:

El número de combinaciones de x elementos en un conjunto de n elementos es dada por:

[tex]C_{n,x} = \frac{n!}{x!(n - x)!}[/tex]

En este problema, combinaciones de 2 elementos de un conjunto de 5, entonces:

[tex]C_{5,2} = \frac{5!}{2!3!} = 10[/tex]

10 partidos distintos se pueden realizar.

Item 5:

El número de arreglos de n elementos viene dado por

[tex]A_n = n![/tex]

En este problema, arreglo de 10 elementos, o sea:

[tex]A_{10} = 10! = 3628800[/tex]

Pueden colocarse de 3,628,800 formas.

Item 6:

El número de arreglos de n elementos, con repeticiones de [tex]n_1, n_2, \cdots n_n[/tex] elementos viene dado por

[tex]A_n^{n_1,n_2,\cdots,n_n} = \frac{n!}{n_1!n_2! \cdots n_n!}[/tex]

En este problema, [tex]n = 9, n_1 = 3, n_2 = 2, n_3 = 4[/tex], por eso:

[tex]A_9^{3,2,4} = \frac{9!}{3!2!4!} = 1260[/tex]

1260 señales distintas pueden indicarse.

Un problema similar es dado en https://brainly.com/question/19022577

Answer:

s=4

Step-by-step explanation:

area of square is side*side

so that means s*s=16

so s = 4

Answer:

  40 in

Step-by-step explanation:

For a width of w, the length is 3w and the area and perimeter are ...

  A = LW = (3w)(w) = 3w^2

  P = 2(L+W) = 2(3w +w) = 8w

We are given the area, so we can find w to be ...

  75 in^2 = 3w^2

  25 in^2 = w^2 . . . . . divide by 3

  5 in = w . . . . . . . . . square root

Then the perimeter is ...

  P = 8w = 8(5 in) = 40 in