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3 years ago

Help solve #2 please!!!

Mathematics

1 answer:

lys-0071 [83]3 years ago

8 0

Answer:

Step-by-step explanation:

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Given f(x)=x-10 . what is f(-7)?

Tomtit [17]

F(-7) = -7 - 10

f(-7) = -17

Hope I helped!

Let me know if you need anything else!

<span>~ Zoe (Rank:'Genius')</span>

8 0

3 years ago

The table and the graph below each show a different relationship between the same two variables, x and y:

gavmur [86]

The value of y would be 20 times more valuable than 12 because the pattern for this table is y=20x. Let me know if I'm right :)

7 0

2 years ago

.........................

malfutka [58]

Pay attention in math class! in higher grade levels, this is gonna show up EVERYDAY! no joke
In these types of problems, you need to use PEMDAS also.

((4)+3)+(13(-2)-5)
(7)+(-31)
7-31
-24
<u><em>C</em></u>

8 0

3 years ago

4/5 ÷ 7/6 + 2/7 ∙2 1/5

Aleksandr [31]

<h2>The value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ $=\dfrac{8}{7}$ </h2>

Step-by-step explanation:

We have,

$\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$

To find, the value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ = ?

∴ $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$

$=\dfrac{4}{5}\times \dfrac{6}{7} +\dfrac{2}{7}.\dfrac{8}{5}$

$=\dfrac{24}{35} +\dfrac{16}{35}$

Taking LCM of denominator, we get

$=\dfrac{24+16}{35}$

$=\dfrac{40}{35}$

Dividing numerator and denominator by 5, we get

$=\dfrac{8}{7}$

∴ The value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ $=\dfrac{8}{7}$

Hence, the value of $\dfrac{4}{5}$ ÷ $\dfrac{7}{6}$ + $\dfrac{2}{7}.2\dfrac{1}{5}$ is equal to $\dfrac{8}{7} .$

4 0

3 years ago

A 2 kg metal cylinder is supplied with 1600J of energy to heat it from 5°C to 13°C. What is the specific heat capacity of the me

Neporo4naja [7]

Answer:

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$

Step-by-step explanation:

According to the First Law of Thermodynamics, the heat received by the metal cylinder is equal to the change in the internal energy. That is:

$Q = \Delta U$

$Q = m \cdot c \cdot \Delta T$

The specific heat is clear in the previous expression and finally computed:

$c = \frac{Q}{m\cdot \Delta T}$

$c = \frac{1600\,J}{(2\,kg)\cdot (13^{\circ}C-5^{\circ}C)}$

$c = 100\,\frac{J}{kg\cdot ^{\circ}C}$

6 0

2 years ago