All categories

3 years ago

1.8 minus 3.7 x = negative 4.2 x + 0.3?

1 answer:

6 0

Combine like terms. First, add 4.2x to both sides. This results in 1.8+0.5x = 0.3, and then 0.5x = 0.3-1.8.

Next, 0.05x = -1.5, or 5x = -150, and x = -30.

Must check this result by substitution of -30 for x in the original equation.

Is 1.8-3.7(-30)=-4.2(-30)+0.3 true or false?

You might be interested in

Solve this equation 11/20 +x= 9/10

Sav [38]

Answer:

7/20

Step-by-step explanation:

$\frac{11}{20}+x=\frac{9}{10} \\x=\frac{7}{20}$

7 0

3 years ago

Read 2 more answers

What is the value of h when the function is converted to vertex form?

elena-s [515]

Answer: h=7

Step-by-step explanation:

By definition the function in vertex form is:

$f(x)=a(x-h)^{2}+k$

Where (h,k) is the vertex of the parabola.

You must group the terms that contain the same variable in the function:

$f(x)=(x^{2}-14x)+29$

Complete the square as following:

$f(x)=(x^{2}-14x+49)+29-49$

Now you must rewrite it as following:

$f(x)=(x-7)^{2}-20$

Then:

h=7

4 0

3 years ago

Read 2 more answers

Order these numbers from least to greatest 4.1072,4.7,4.608,4.67

ira [324]

4.7,4.67,4.67,4.1072

6 0

4 years ago

Use the Fundamental Theorem of Calculus to find the area of the region between the graph of the function x5 + 8x4 + 2x2 + 5x + 1

belka [17]

Answer:

The area of the region between the graph of the given function and the x-axis = 25,351 units²

Step-by-step explanation:

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15

If 'f' is a continuous on [a ,b] then the function

$F(x) = \int\limits^a_b {f(x)} \, dx$

By using integration formula

$\int{x^n} \, dx = \frac{x^{n+1} }{n+1} +c$

Given x⁵ + 8 x⁴ + 2 x² + 5 x + 15 in the interval [-6,6]

$\int\limits^6_^-6} (x^{5} + 8 x^{4} + 2 x^{2} + 5 x + 15) )dx$

On integration , we get

= $(\frac{x^{6} }{6} + \frac{8 x^{5} }{5} + 2 \frac{x^{3} }{3} +\frac{5 x^{2} }{2} + 15 x)^{6} _{-6}$

$F(x) = \int\limits^a_b {f(x)} \, dx = F(b) -F(a)$

= $(\frac{6^{6} }{6} + \frac{8 6^{5} }{5} + 2 \frac{6^{3} }{3} +\frac{5 6^{2} }{2} + 15X 6) - ((\frac{(-6)^{6} }{6} + \frac{8 (-6)^{5} }{5} + 2 \frac{(-6)^{3} }{3} +\frac{5 (-6)^{2} }{2} + 15 (-6))$