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3 years ago

Find the mistake in the number 456,789 word form version four hundred fifty six thousand and seven hundred eighty nine

Mathematics

1 answer:

tatiyna3 years ago

4 0

Answer:

456,789

Step-by-step explanation: Four Hundred Fifty Six Thousand And Seven Hundred Eighty Nine.

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The Johnson family want to start a college fund for their daughter Gabriella. They put $63, 000 in to an account that grows at a

marissa [1.9K]

Answer:

34.13 years

Step-by-step explanation:

Given formula:

G(t)= 63000(1 + 0.0255/4)^4t

Need to find the value of t when G(t) = 150000

150000 = 63000( 1+ 0.0255/4)^4t
( 1+ 0.0255/4)^4t = 150000/63000
( 1+ 0.0255/4)^4t = 2.3809
4t* log( 1+ 0.0255/4) = log 2.3809
4t *0.00275983965 = 0.37674115501
t = 0.37674115501/(4*0.00275983965)
t = 34.13 years rounded

6 0

3 years ago

Acute △ABC with angles α, β, and γ is inscribed in a circle. Tangents to the circle at points A, B, and C intersect in points M,

SOVA2 [1]

Answer:

The measures of angles of triangle MNP are

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Find the measure of arcs AB, BC and AC

we know that

The inscribed angle is half that of the arc it comprises.

$\gamma=\frac{1}{2}[arc\ AB] ----> arc\ AB=2\gamma\\\alpha=\frac{1}{2}[arc\ BC] ----> arc\ BC=2\alpha\\\beta=\frac{1}{2}[arc\ AC] ----> arc\ AC=2\beta$

step 2

Find the measure of angle M

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

$M=\frac{1}{2}[arc\ AB+arc\ BC-arc\ AC]$

substitute

$M=\frac{1}{2}[2\gamma+2\alpha-2\beta]\\M=[\gamma+\alpha-\beta]$

step 3

Find the measure of angle N

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

$N=\frac{1}{2}[arc\ AC+arc\ BC-arc\ AB]$

substitute

$N=\frac{1}{2}[2\beta+2\alpha-2\gamma]\\N=[\beta+\alpha-\gamma]$

step 4

Find the measure of angle P

we know that

The measurement of the outer angle is the semi-difference of the arcs it encompasses.

$P=\frac{1}{2}[arc\ AC+arc\ AB-arc\ BC]$

substitute

$P=\frac{1}{2}[2\beta+2\gamma-2\alpha]\\P=[\beta+\gamma-\alpha]$

5 0

3 years ago

I would love for someone to help me out :)

Kryger [21]

We can solve this problem in different ways:

0. Statement: As angle 1 and 3 are vertical angles, and angle 1, 5, 4 form a straigth angle, so angle 2 and (algle 4 + angle 5) are vertcal angles too.

Reason:

$\begin{gathered} \angle1+\angle4+\angle5=180=\angle2+\angle3(share\text{ the same line)} \\ \angle1=\angle3(verticl\text{ angles)} \\ So, \\ \angle4+\angle5=\angle2 \end{gathered}$

3 0

1 year ago

Calcula la altura de un cono se 235.62cm³ de volumen si el radio de la base mide 5cm

Ber [7]

Answer:

Datos: r= 24cm. g= 74 cm. ...

- La altura h.

- La generatriz.

- el radio. Que entre ellos forman un triangulo rectángulo siendo la generatriz la hipotenusa y el radio la base, entonces: ...

g²=r²+h² despejando la altura h se tiene que:

h=√g²-r² h= √ (74cm)² - (24 cm)²

es. un paso para que. puedas resolver

6 0

3 years ago

Find the lcm of ÷ 16a^4+4a^2+169,8a^3+2a(2a+13)+16a^2

klio [65]

Answer:

Step-by-step explanation:

Hope this helps u !!

5 0

3 years ago