All categories

3 years ago

Write a number sentence that does not involve regrouping using this sum and two 2- digit addends

Mathematics

1 answer:

melisa1 [442]3 years ago

5 0

Using the sum (addition) and two 2-digit addends, 12 + 34 can be answered without regrouping.

You might be interested in

Graph and label the image of the figure below after a dilation by a factor of 1/2.

neonofarm [45]

Answer:

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

see graph below

Explanation:

Given:

The image of a quadrilateral on a coordinate plane

To find:

The coordinates of the new image after dilation of 1/2 have been applied to the original image.

Then graph the coordinates

First, we need to state the coordinates of the original image:

M = (3, -2)

F = (4, -2)

L = (1, -5)

W = (5, -5)

Next, we will apply a scale factor of 1/2:

$\begin{gathered} Dilation\text{ rule:} \\ (x,\text{ y\rparen}\rightarrow(kx,\text{ ky\rparen} \\ where\text{ k = scale factor} \\ \\ scale\text{ factor = 1/2} \\ M^{\prime}\text{ = \lparen}\frac{1}{2}(3),\text{ }\frac{1}{2}(-2)) \\ M^{\prime}\text{ = \lparen}\frac{3}{2},\text{ -1\rparen} \\ \\ F\text{ = \lparen}\frac{1}{2}(4),\text{ }\frac{1}{2}(-2)) \\ F^{\prime}\text{ = \lparen2, -1\rparen} \end{gathered}$ $\begin{gathered} L\text{ = \lparen}\frac{1}{2}(1),\text{ }\frac{1}{2}(-5)) \\ L^{\prime}\text{ = \lparen}\frac{1}{2},\text{ }\frac{-5}{2}) \\ \\ W\text{ = \lparen}\frac{1}{2}(5),\text{ }\frac{1}{2}(-5)) \\ W^{\prime}\text{ = \lparen}\frac{5}{2},\text{ }\frac{-5}{2}) \end{gathered}$

The new coordinates:

M' (3/2, -1), F' (2, -1), L' (1/2, -5/2), W' (5/2, -5/2)

M' (1.5, -1), F' (2, -1), L' (0.5 -2.5), W' (2.5, -2.5)

Plotting the coordinates:

3 0

9 months ago

You have decided to invest $1000 in a savings bond that pays 4% interest, compounded semi-annually. What will the bond be worth

Evgen [1.6K]

Answer:

$2191.12

Step-by-step explanation:

We are asked to find the value of a bond after 10 years, if you invest $1000 in a savings bond that pays 4% interest, compounded semi-annually.

$FV=C_0\times (1+r)^n$ , where,

$C_0=\text{Initial amount}$ ,

r = Rate of return in decimal form.

n = Number of periods.

Since interest is compounded semi-annually, so 'n' will be 2 times 10 that is 20.

$4\%=\frac{4}{100}=0.04$

$FV=\$1,000\times (1+0.04)^{20}$

$FV=\$1,000\times (1.04)^{20}$

$FV=\$1,000\times 2.1911231430334194$

$FV=\$2191.1231430334194$

$FV\approx \$2191.12$

Therefore, the bond would be $2191.12 worth in 10 years.

3 0

3 years ago

With Obesity on the rise, a Doctor wants to see if there is a linear relationship between the Age and Weight and estimating a pe

Romashka [77]

Answer:

112.569 ( D )

Step-by-step explanation:

Applying the estimated Regression Equation

y = b1X1 + b2X2 + a

b1 = ((SPX1Y)*(SSX2)-(SPX1X2)*(SPX2Y)) / ((SSX1)*(SSX2)-(SPX1X2)*(SPX1X2)) = 596494.5/635355.88 = 0.93884

b2 = ((SPX2Y)*(SSX1)-(SPX1X2)*(SPX1Y)) / ((SSX1)*(SSX2)-(SPX1X2)*(SPX1X2)) = 196481.5/635355.88 = 0.30925

a = MY - b1MX1 - b2MX2 = 149.25 - (0.94*61.31) - (0.31*193.88) = 31.73252

y = 0.939X1 + 0.309X2 + 31.733

For x1 ( age ) =39, and x2(weight) =143

y = (0.93884*39) + (0.30925*143) + 31.73252= 112.569

where

Sum of X1 = 981

Sum of X2 = 3102

Sum of Y = 2388

Mean X1 = 61.3125

Mean X2 = 193.875

Mean Y = 149.25

attached is the Tabular calculation of the required values needed for estimated regression equation

8 0

2 years ago

A solution initially contains 200 bacteria. 1. Assuming the number y increases at a rate proportional to the number present, wri

GuDViN [60]

Answer:

1. $\frac{dy}{dt}=ky$

2.543.6

Step-by-step explanation:

We are given that

y(0)=200

Let y be the number of bacteria at any time

$\frac{dy}{dt}$ =Number of bacteria per unit time

$\frac{dy}{dt}\proportional y$

$\frac{dy}{dt}=ky$

Where k=Proportionality constant

2. $\frac{dy}{y}=kdt$ ,y'(0)=100

Integrating on both sides then, we get

$lny=kt+C$

We have y(0)=200

Substitute the values then , we get

$ln 200=k(0)+C$

$C=ln 200$

Substitute the value of C then we get

$ln y=kt+ln 200$

$ln y-ln200=kt$

$ln\frac{y}{200}=kt$

$\frac{y}{200}=e^{kt}$

$y=200e^{kt}$

Differentiate w.r.t

$y'=200ke^{kt}$

Substitute the given condition then, we get

$100=200ke^{0}=200 \;because \;e^0=1$

$k=\frac{100}{200}=\frac{1}{2}$

$y=200e^{\frac{t}{2}}$

Substitute t=2

Then, we get $y=200e^{\frac{2}{2}}=200e$

$y=200(2.718)=543.6=543.6$

e=2.718

Hence, the number of bacteria after 2 hours=543.6

4 0

3 years ago

The value of the fraction 3c+2/4 one less than the value of the fraction 5c-1/3.

Viefleur [7K]

Answer:

7/8 > 5/6

Step-by-step explanation:

A) 7/8

We can compare this as follows.

Lets say both are equal.

Cross multiplying these we get 40=42

We get 40<42. In fraction we get

In case if you want to convert this to decimal, we get;

5/6 = 0.833 and 7/8 = 0.875

We get 5/6<7/8

B) 4/5

Similarly we get 4/5 = 0.8 and 5/6 = 0.833

Here 4/5<5/6

C) 3/4

we get 3/4 = 0.75 and 5/6 = 0.833

3/4<5/6

D) 2/3

we get 2/3 = 0.66 and 5/6 = 0.833

2/3<5/6

Step-by-step explanation:

5 0

2 years ago