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3 years ago

Find all the values of n that make the following statement true :|n| equel to 23

Mathematics

1 answer:

omeli [17]3 years ago

7 0

Nothing multiplies into the number 23

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Solve the word problem. The height of a triangle with base b and area A is given by the formula h = 2A/B . What is the height

jeka94

$h = \frac{2A}{b} \ \ \ and\ \ \ A=26\ cm^2\ \ \ and\ \ \ b=8\ cm\\\\h= \frac{2\cdot26}{8} = \frac{2\cdot2\cdot13}{2\cdot2\cdot2} = \frac{13}{2} =6.5\ [cm]\\\\Ans.\ D$

4 0

3 years ago

The length of a rectangle is 3 cm less than twice its width if the perimeter of the rectangle is 18 cm find its dimensions

mestny [16]

Answer:

let width be x

Step-by-step explanation:

2x-6+3+2x-6+3=18

4x+6=18

4x=18-6

4x=12

x=12/4

x=3

3 0

3 years ago

Zoom in to see better

KonstantinChe [14]

Answer:

1:Gavin is incorrect.

2:4

3: 1/4

4: 1/12

Step-by-step explanation:

5 0

3 years ago

VT is the diagonal of a rhombus. What is the slope of the other diagonal?

noname [10]

Answer:

Step-by-step explanation:

Remark

The slope of the given segment is found by using the 2 end points.

V(3,2) and T(5,7)

Slope of VT

Givens

y2 = 7
y1 = 2
x2 = 5
x1 = 3

Formula

m = (y2 - y1)/(x2 - x1)

Solution

m = (7 - 2)/(5 -3)

m = 5/2

Slope of other diagonal

m * m1 = - 1

5/2 * m1 = - 1

5m1 = -1*2 = -2

m1 = -2/5

6 0

3 years ago

The equation y = \large 1\frac{1}{2}x represents the number of cups of dried fruit, y, needed to make x pounds of granola. Deter

Natasha2012 [34]

Answer:

$(1\frac{1}{2},1)$ - False

$(4,6)$ - True

$(18,12)$ -- False

$(0,0)$ -- True

$(2\frac{1}{2},3\frac{3}{4})$ -- True

Step-by-step explanation:

The points are

$(1\frac{1}{2},1)$ , $(4,6)$ , $(18,12)$ , $(0,0)$ and $(2\frac{1}{2},3\frac{3}{4})$ ---- missing from the question

Given

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

Required

Determine if each of the points would be on $y = 1\frac{1}{2}x$

To do this, we simply substitute the value of x and of each point in $y = 1\frac{1}{2}x$ .

(a) $(1\frac{1}{2},1)$

In this case;

$x = 1\frac{1}{2}$ and $y = 1$

$y = 1\frac{1}{2}x$ becomes

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

$y = \frac{3}{2} * \frac{3}{2}$

$y = \frac{9}{4}$

$y = 2\frac{1}{4}$

The point $(1\frac{1}{2},1)$ won't be on the graph because the corresponding value of y for $x = 1\frac{1}{2}$ is $y = 2\frac{1}{4}$

(b) $(4,6)$

In this case;

$x = 4$

$y = 6$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 4$

$y = \frac{3}{2} * 4$

$y = \frac{3* 4}{2}$

$y = \frac{12}{2}$

$y = 6$

The point $(4,6)$ would be on the graph because the corresponding value of y for $x = 4$ is $y = 6$

(c) $(18,12)$

In this case:

$x = 18;y = 12$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 18$

$y = \frac{3}{2} * 18$

$y = \frac{3* 18}{2}$

$y = \frac{54}{2}$

$y = 27$

The point $(18,12)$ wouldn't be on the graph because the corresponding value of y for $x = 18$ is $y = 12$

(d) $(0,0)$

In this case;

$x =0; y = 0$

$y = 1\frac{1}{2}x$ becomes

$y = 1\frac{1}{2} * 0$

$y = 0$

The point $(0,0)$ would be on the graph because the corresponding value of y for $x = 0$ is $y = 0$

(e) $(2\frac{1}{2},3\frac{3}{4})$

In this case:

$x = 2\frac{1}{2}$ ; $y = 3\frac{3}{4}$

$y = 1\frac{1}{2}x$ becomes

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

$y = \frac{3}{2} * \frac{5}{2}$

$y = \frac{15}{4}$

$y = 3\frac{3}{4}$

The point $(2\frac{1}{2},3\frac{3}{4})$ would be on the graph because the corresponding value of y for $x = 2\frac{1}{2}$ is $y = 3\frac{3}{4}$

3 0

3 years ago