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2 years ago

Where do the graphs Of linear equations 2x+3y=4 and 5x+6y=7 intersect

1 answer:

4 0

5x + 6y = 7 SUBSTITUTION METHOD
2x + 2y = 4

5x + 6y = 7 Solve for either x or y in 1 of the 2 equations.
- 6y - 6y
-----------------------
5x = -6y + 7
----- ------ ----
5 5 5

x = -6/5y + 7/5

2(-6/5y + 7/5) + 2y = 4 Substitute the expression you got for x or y into
-12/5y + 14/5 + 2y = 4 the other equation.
-2/5y + 14/5 = 4
- 14/5 - 14/5
------------------------------
-2/5y = 6/5
-------- --------
- 2/5y - 2/5y

y = -3

5x + 6(-3) = 7 Plug in the value you got for y in step 2 into the
5x - 18 = 7 equation that you didn't use.
+ 18 + 18
------------------
5x = 25
--- -----
5 5

x = 5

(x,y) ⇒ coordinate
(5,-3) ⇒ final answer

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3x+5=5x-3

madam [21]

answer= d Step-by-step explanation:

6 0

2 years ago

DG¯¯¯¯¯¯ , EG¯¯¯¯¯ , and FG¯¯¯¯¯ are perpendicular bisectors of the sides of △ABC . DG=5 cm and BD=12 cm.

jonny [76]

Given that DG, EG, and FG are perpendicular bisectors of the sides of △ABC, this means that the point of intersection, G, is the circumcenter of the triangle and hence AG, BG, and CG are equal.

Given that DG = 5 cm and BD = 12 cm, then

$BG= \sqrt{DG^2+BD^2} \\ \\ = \sqrt{5^2+12^2} = \sqrt{25+144} \\ \\ = \sqrt{169} =13$

Since AG = BG = CG, therefore, CG = 13 cm.

3 0

2 years ago

A researcher computes the following 2 x 3 between-subjects ANOVA, in which 11 participants were observed in each group. Which ef

pishuonlain [190]

Answer:

D. There were no significant effects.

Step-by-step explanation:

The table below shows the representation of the significance level using the two-way between subjects ANOVA.

Source of Variation SS df MS F P-value

Factor A 10 1 10 0.21 0.660

Factor B 50 2 25 0.52 0.6235

A × B 40 2 20 0.42 0.6783

Error 240 5 48 - -

Total 340 10 - - -

From the table above , the SS(B) is determined as follows:

SS(B) = SS(Total)-SS(Error-(A×B)-A)

= 340-(240-40-10)

= 50

A researcher computes the following 2 x 3 between-subjects ANOVA;

k=2

n=3

N(total) = no of participants observed in each group =11

df for Factor A= (k-1)

=(2-1)

df for Factor B = (n-1)

=(3-1)

df for A × B

= 2 × 1

= 2

df factor for total

=(N-1)

=11-1

=10

MS = SS/df

Thus, from the table, the P-Value for all data is greater than 0.05, therefore we fail to reject H₀.

4 0

2 years ago

The number of bacteria after t hours is given by N(t)=250 e^0.15t a) Find the initial number of bacteria and the rate of growth

Art [367]

Answer:

a) $N_0=250\; k=0.15$

b) 334,858 bacteria

c) 4.67 hours

d) 2 hours

Step-by-step explanation:

a) Initial number of bacteria is the coefficient, that is, 250. And the growth rate is the coefficient besides “t”: 0.15. It’s rate of growth because of its positive sign; when it’s negative, it’s taken as rate of decay.

Another way to see that is the following:

Initial number of bacteria is N(0), which implies $t=0$ . And $N(0)=N_0$ . The process is:

$N(t)=250 e^{0.15t}\\N(0)=250 e^{0.15(0)}\\ N_0=250e^{0}\\N_0=250\cdot1\\ N_0=250$

b) After 2 days means $t=48$ . So, we just replace and operate:

$N(t)=250 e^{0.15t}\\N(48)=250 e^{0.15(48)}\\ N(48)=250e^{7.2}\\N(48)=334,858\;\text{bacteria}$

c) $N(t_1)=4000; \;t_1=?$

$N(t)=250 e^{0.15t}\\4000=250 e^{0.15t_1}\\ \dfrac{4000}{250}= e^{0.15t_1}\\16= e^{0.15t_1}\\ \ln{16}= \ln{e^{0.15t_1}} \\ \ln{16}=0.15t_1 \\ \dfrac{\ln{16}}{0.15}=t_1=4.67\approx 5\;h$

d) $t_2=?\; (N_0→3N_0 \Longrightarrow 250 → 3\cdot250 =750)$

$N(t)=250 e^{0.15t}\\ 750=250 e^{0.15t_2} \\ \ln{3} =\ln{e^{0.15t_2}}\\ t_2=\dfrac{\ln{3}}{0.15} = 2.99 \approx 3\;h$

6 0

3 years ago

How many times can 6 go into 5920

USPshnik [31]

That answer would be: 987.6666..... (6 repeating)

8 0

2 years ago

Read 2 more answers