All categories

3 years ago

WILL GIVE BRAINLIEST!!!! LOOK BELOW TO ANSWER MY QUESTION.

Mathematics

1 answer:

galben [10]3 years ago

7 0

1.20(35)

this would be the same as :
(1 + 0.2)* 35 <==
or it also could have been 35 + 0.20(35)

You might be interested in

The graph of h(x) is shown. Graph of h of x that begins in quadrant two and decreases rapidly following the vertical line which

fgiga [73]

Answer:

x-intercept

The point at which the curve crosses the x-axis, so when y = 0.

From inspection of the graph, the curve appears to cross the x-axis when x = -4, so the x-intercept is (-4, 0)

y-intercept

The point at which the curve crosses the y-axis, so when x = 0.

From inspection of the graph, the curve appears to cross the y-axis when y = -1, so the y-intercept is (0, -1)

Asymptote

A line which the curve gets infinitely close to, but never touches.

From inspection of the graph, the curve appears to get infinitely close to but never touches the vertical line at x = -5, so the vertical asymptote is x = -5

(Please note: we cannot be sure that there is a horizontal asymptote at y = -2 without knowing the equation of the graph, or seeing a larger portion of the graph).

5 0

2 years ago

Read 2 more answers

Let a triangle with integer angle measures have one angle of 116°. What is the largest possible angle measure opposite the small

sveta [45]

Answer:

what will I do here?

Step-by-step explanation:

sorry im still grade 3

3 0

2 years ago

A fraction becomes 4÷5 if 1 is added to both numerator and denominator. If, however, 5 is subtracted from both numerator and den

tia_tia [17]

Answer:

$\dfrac{7}{9}$

Step-by-step explanation:

$\dfrac{x+1}{y+1}=\dfrac{4}{5}\\\Rightarrow 5x-4y=-1$

$\dfrac{x-5}{y-5}=\dfrac{1}{2}\\\Rightarrow 2x-y=5$

Putting it in matrix form

$\begin{bmatrix}a_{1}&b_{1}\\a_{2}&b_{2}\end{bmatrix}{\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}{c_{1}}\\{c_{2}}\end{bmatrix}\\\Rightarrow\begin{bmatrix}5 & -4\\2 & -1\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}-1\\ 5\end{bmatrix}$

From Cramer's rule we have

$x=\dfrac{\begin{vmatrix}c_1 &b_1 \\ c_2 & b_2\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow x=\dfrac{\begin{vmatrix}-1 &-4 \\ 5 & -1\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1\end{vmatrix}}\\\Rightarrow x=\dfrac{1+20}{-5+8}\\\Rightarrow x=7$

$y=\dfrac{\begin{vmatrix}a_1 &c_1 \\ a_2 & c_1\end{vmatrix}}{\begin{vmatrix}a_1 &b_1 \\ a_2& b_2\end{vmatrix}}\\\Rightarrow y=\dfrac{\begin{vmatrix}5 &-1 \\ 2 & 5\end{vmatrix}}{\begin{vmatrix}5 &-4 \\ 2& -1 \end{vmatrix}}\\\Rightarrow y=\dfrac{25+2}{-5+8}\\\Rightarrow y=9$

Verifying the results

$\dfrac{7+1}{9+1}=\dfrac{8}{10}=\dfrac{4}{5}$

$\dfrac{7-5}{9-5}=\dfrac{2}{4}=\dfrac{1}{2}$

Hence, the fraction is $\dfrac{7}{9}$ .

6 0

2 years ago

The table shows scores that eight students earned on four different quizzes use scatter plots to solve

Vlad [161]

The answer is either B or D but I believe it is B

7 0

3 years ago

A country's population in 1993 was 127 million. In 2000 it was 132 million. Estimate the population in 2008 using the exponentia

12345 [234]

To solve the exponential growth application question we proceed as follows;
suppose the time, t between 1993 to 2000 is such that in 1993, t=0 and in 2000, t=7.
Note theta the population is in millions;
The exponential formula is given by:
f(t)=ae^(kt)
where;
f(t) =current value
a=initial value
k=constant of proportionality
t=time
substituting the values we have in our formula we get:
132=127e^(7k)
132/127=e^(7k)
introducing the natural logs we get:
ln (132/127)=7k
k=[ln(132/127)]/7
k=0.0055
Thus our formula will be:
f(t)=127e^(0.0055t)
The population in 2008 will be:
f(t)=127e^(15*0.0055)=127e^(0.0825)=137.922
Thus the population in 2008 is appropriately 138 million.

5 0

3 years ago