CAN SOMEONE PLEASE HELP WITH THIS GEOMETRY QUESTION ASAP!!! And both of those are 10 degrees

If the original price is $40.00 and the percent of markup is 8% what is the value of the markup ?

Ne4ueva [31]

Answer=$3.20

8%=0.08

$40.00*0.08=$3.20

The markup was $3.20

7 0

3 years ago

For a school fund raising project Aiden and Jacob collected money in the ratio 3.5 to 6.8 Jacob collected $115.50 more than Aide

Liula [17]

FOR A SCHOOOL NOW FIRST IF THE Ratio was 3.5 to 6.8 then Jacob Collected 115.50 it would be 207.56

7 0

3 years ago

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Carla trabaja en el área de compras de una empresa que tiene 8 sucursales a lo largo de Chile. A pedido de su jefe necesita comp

Anvisha [2.4K]

Answer:

Los gastos totales en los que incurre Carla = (36x + 28y + 3z) pesos

Step-by-step explanation:

A partir de la pregunta, a continuación se proporciona la cantidad de elementos necesarios:

21 lápices, 8 carpetas, 3 broches, 10 resaltadores, 6 blocs de notas adhesivas y 9 rotuladores.

Los precios de cada artículo se dan a continuación

Precio de cada lápiz = x pesos

Precio de cada carpeta = y pesos

Precio de cada broche = z pesos

El precio de cada resaltador es el doble que el de cada carpeta

= 2 × y pesos

= 2y pesos

El precio de cada barra de notas adhesivas es el doble del precio de cada lápiz

= 2 × x pesos

= 2x pesos

El precio de cada marcador (bolígrafo) es el triple del precio de cada lápiz.

= 3 × x pesos

= 3x pesos

Indique el gasto total en el que incurrirá Carla.

Los gastos totales en los que incurrirá Carla =

(21 × x pesos) + (8 × y pesos) + (3 × z pesos) + (10 × 2y pesos) + (6 × x pesos) + (9 × x pesos)

= (21x + 8y + 3z + 20y + 6x + 9x) pesos

Recopilar términos similares

= (21x + 6x + 9x + 8y + 20y + 3z) pesos

= (36x + 28y + 3z) pesos

Por lo tanto, los gastos totales en los que incurre Carla = (36x + 28y + 3z) pesos

6 0

3 years ago

The area of a rectangular field is 48m2 and one of its sides is 6m . How long will a lady take to cross the field diagonally at

tatiyna

Answer:

It will take her approximately 28 seconds to cross the diagonal of the pitch

Step-by-step explanation:

If we can find the length of the diameter of the field (in Metres), we can easily find the time it takes her to cross this length using motion equations.

Step one: Find the length of the field from its area.

The field is rectangular in shape. The area of the field is 48m^2. To get the area of the field, we had to multiply its length and its breadth.

i.e area = length X breadth.

Let us assume that the breadth = 6m

its length will be area / breadth = 42/ 6 = 7m

The length of the pitch = 7 m

Step Two: Find the length of the diagonal using Pythagoras' theorem.

From Pythagoras'theorem, the length of the diagonal can be obtained by

$diagonal = \sqrt{l^2 +b^2}$

$diagonal = \sqrt{6^2+ 7^2} = 9.22m$

The length of the diagonal is 9.22m

Step three: Find the time using her speed and the distance she needs to cover.

Time = distance / speed

time = 9.22 /20 = 0.46 minutes

It will take her approximately 28 seconds to cross the diagonal of the pitch

6 0

3 years ago

Calc help pretty please

Gre4nikov [31]

Answer:

A. 77.4

Step-by-step explanation:

Linear approximation formula

$L(x)=f(a)+f'(a)(x-a)$

Given function:

$f(x)=(x+1)^2$

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times \dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\boxed{\begin{minipage}{5 cm}\underline{Differentiating $x^n$}\\\\If $y=x^n$, then $\dfrac{\text{d}y}{\text{d}x}=xn^{n-1}$\\\end{minipage}}$

$\boxed{\begin{minipage}{4 cm}\underline{Differentiating $ax$}\\\\If $y=ax$, then $\dfrac{\text{d}y}{\text{d}x}=a$\\\end{minipage}}$

Use the chain rule to differentiate the function.

$\textsf{Let }\:y=u^2\:\textsf{ where }u=(x+1)$

Differentiate the two parts separately:

$y=u^2 \implies \dfrac{\text{d}y}{\text{d}u}=2u$

$u=x+1 \implies \dfrac{\text{d}u}{\text{d}x}=1$

Put everything back into the chain rule formula:

$\begin{aligned} \implies \dfrac{\text{d}y}{\text{d}x} & =2u \times 1\\ & = 2u \\ & = 2(x+1)\\ & = 2x+2 \end{aligned}$

$\textsf{Therefore, }\:f'(x)=2x+2.$ .

The linear approximation at a = 8 is:

$\begin{aligned}L(x) & =f(a)+f'(a)(x-a)\\\\\implies L(x) & = f(8)+f'(8)(x-8)\\& = (8+1)^2+(2(8)+2)(x-8)\\& = 81+18(x-8)\\& = 18x-63\end{aligned}$

Finally, substitute x = 7.8 into the linear approximation equation:

$\begin{aligned}\implies L(7.8) & =18(7.8)-63\\& = 140.4-63\\& = 77.4\end{aligned}$

4 0

2 years ago

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