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3 years ago

What's the simplified form of 2x + 3 – X + 5? A.X-2 B.X-8 C. x+8 D. 3x + 8

Mathematics

1 answer:

allochka39001 [22]3 years ago

8 0

Answer:

Step-by-step explanation:

Group like terms

= 2x - x +3 +5

Add the similar 'elements'

= x + 3 + 5

Add the numbers

3 + 5 = 8 + x

= x+8

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3 years ago

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I need help please.

sweet-ann [11.9K]

Answer:

Third Option: $y = \frac{5}{4}x$

Step-by-step explanation:

Given the points on the graph, (4, 5) and (-4, -5):

In order to determine the equation of the given graph in slope-intercept form, y = mx + b:

Use the given points to solve for the slope:

Let (x₁, y₁) = (-4, -5)

(x₂, y₂) = (4, 5)

m = (y₂ - y₁)/(x₂ - x₁)

$m = \frac{5 - (5)}{4 - (-4)} = \frac{5 + 5}{4 + 4} = \frac{10}{8} = \frac{5}{4}$

Therefore, the slope of the line is: $m = \frac{5}{4}$ .

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y = mx + b

5 = $\frac{5}{4} (4)$ + b

5 = 5 + b

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0 = b

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2 years ago

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3 years ago

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Solve the system: 2/x - 3/y =-5; 4/x + 6\y =14

sertanlavr [38]

The solution is $x = 2 \text{ and } y = \frac{1}{2}$

Solution:

Let us assume,

$a = \frac{1}{x}\\\\b = \frac{1}{y}$

Given system of equations are:

$\frac{2}{x} - \frac{3}{y} = -5$

$\frac{4}{x} + \frac{6}{y} = 14$

Rewrite the equation using "a" and "b"

2a - 3b = -5 ------------ eqn 1

4a + 6b = 14 -------------- eqn 2

Let us solve eqn 1 and eqn 2

Multiply eqn 1 by 2

2(2a - 3b = -5)

4a - 6b = -10 ------------- eqn 3

Add eqn 2 and eqn 3

4a + 6b = 14

4a - 6b = -10

( - ) --------------------

8a = 4

$a = \frac{4}{8}\\\\a = \frac{1}{2}$

Substitute a = 1/2 in eqn 1

$2(\frac{1}{2}) -3b = -5\\\\1 - 3b = -5\\\\3b = 6\\\\b = 2$

Now let us go back to our assumed values

Substitute a = 1/2 in assumed values

$a = \frac{1}{x}\\\\\frac{1}{2} = \frac{1}{x}\\\\x = 2$

Substitute b = 2 in assumed value

$b = \frac{1}{y}\\\\2 = \frac{1}{y}\\\\y = \frac{1}{2}$

Thus the solution is $x = 2 \text{ and } y = \frac{1}{2}$

3 0

2 years ago

An article reports the following data on yield (y), mean temperature over the period between date of coming into hops and date o

skelet666 [1.2K]

Answer:

x1=c(16.7,17.4,18.4,16.8,18.9,17.1,17.3,18.2,21.3,21.2,20.7,18.5)

x2=c(30,42,47,47,43,41,48,44,43,50,56,60)

y=c(210,110,103,103,91,76,73,70,68,53,45,31)

mod=lm(y~x1+x2)

summary(mod)

R output: Call:

lm(formula = y ~ x1 + x2)

Residuals:

Min 1Q Median 3Q Max

-41.730 -12.174 0.791 12.374 40.093

Coefficients:

Estimate Std. Error t value Pr(>|t|)

(Intercept) 415.113 82.517 5.031 0.000709 ***

x1 -6.593 4.859 -1.357 0.207913

x2 -4.504 1.071 -4.204 0.002292 **

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 24.45 on 9 degrees of freedom

Multiple R-squared: 0.768, Adjusted R-squared: 0.7164

F-statistic: 14.9 on 2 and 9 DF, p-value: 0.001395

a). y=415.113 +(-6.593)x1 +(-4.504)x2

b). s=24.45

c). y =415.113 +(-6.593)*21.3 +(-4.504)*43 =81.0101

residual =68-81.0101 = -13.0101

d). F=14.9

P=0.0014

There is convincing evidence at least one of the explanatory variables is significant predictor of the response.

e). newdata=data.frame(x1=21.3, x2=43)

# confidence interval

predict(mod, newdata, interval="confidence")

#prediction interval

predict(mod, newdata, interval="predict")

confidence interval

> predict(mod, newdata, interval="confidence",level=.95)

fit lwr upr

1 81.03364 43.52379 118.5435

95% CI = (43.52, 118.54)

f). #prediction interval

> predict(mod, newdata, interval="predict",level=.95)

fit lwr upr

1 81.03364 14.19586 147.8714

95% PI=(14.20, 147.87)

g). No, there is not evidence this factor is significant. It should be dropped from the model.

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3 years ago